New Methods in ParameterizedAlgorithms and Complexity

Daniel Lokshtanov

Dissertation for the degree of Philosophiae Doctor (PhD)

University of Bergen

Norway

April 2009

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Acknowledgements

First of all, I would like to thank my supervisor Pinar Heggernes for her help andguidance, both with my reasearch and with all other aspects of being a graduatestudent. This thesis would not have existed without her.

Already in advance, I would like to thank my PhD committee, Maria Chudnovsky,Rolf Niedermeier and Sverre Storøy.

I also want to thank all my co-authors, Noga Alon, Oren Ben-Zwi, Hans Bod-laender, Michael Dom, Michael Fellows, Henning Fernau, Fedor V. Fomin, PetrA. Golovach, Fabrizio Grandoni, Pinar Heggernes, Danny Hermelin, Jan Kra-tochvıl, Elena Losievskaja, Federico Mancini, Rodica Mihai, Neeldhara Misra,Matthias Mnich, Ilan Newman, Charis Papadopoulos, Eelko Penninkx, DanielRaible, Venkatesh Raman, Frances Rosamond, Saket Saurabh, Somnath Sikdar,Christian Sloper, Stefan Szeider, Jan Arne Telle, Dimitrios Thilikos, CarstenThomassen and Yngve Villanger. A special thanks goes to Saket Saurabh, withwhom I have shared many sleepless nights and cups of coffee preparing this work.

Thanks to my family, especially to my mother and father. In more than one way,i am here largely because of you.

I would also like to thank my friends for filling my life with things other thandiscrete mathematics. You mean a lot to me, hugs to all of you and kisses to onein particular.

Thanks also to the Algorithms group at the University of Bergen, you are a greatlot. An extra thanks goes to my table-tennis opponents, even though this thesisexists despite of your efforts to keep me away from it.

Finally, thanks to Pasha for being furry and having whiskers.

Bergen, April 2009Daniel Lokshtanov

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Contents

1 Introduction 1

1.1 Overview of the Thesis . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2 Notation and Definitions . . . . . . . . . . . . . . . . . . . . . . . 3

I Overview of the Field 11

2 Algorithmic Techniques 13

2.1 Bounded Search Trees . . . . . . . . . . . . . . . . . . . . . . . . 13

2.2 Reduction Rules and Branching Recurrences . . . . . . . . . . . . 14

2.3 Iterative Compression . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.4 Dynamic Programming and Courcelle’s Theorem . . . . . . . . . . 19

2.5 Well Quasi Ordering . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.6 Win / Wins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.6.1 Bidimensionality . . . . . . . . . . . . . . . . . . . . . . . 23

2.7 Color Coding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.8 Integer Linear Programming . . . . . . . . . . . . . . . . . . . . . 26

3 Running Time Lower Bounds 31

3.1 NP-hardness vs. ParaNP-hardness . . . . . . . . . . . . . . . . . 31

3.2 Hardness for W[1] . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

3.3 Hardness for W[2] . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

4 Kernelization 37

4.1 Reduction Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

4.2 Crown Reductions . . . . . . . . . . . . . . . . . . . . . . . . . . 39

4.3 LP based kernelization . . . . . . . . . . . . . . . . . . . . . . . . 41

5 Kernelization Lower Bounds 43

5.1 Composition Algorithms . . . . . . . . . . . . . . . . . . . . . . . 43

5.2 Polynomial Parameter Transformations . . . . . . . . . . . . . . . 46

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II New Methods 49

6 Algorithmic Method: Chromatic Coding 516.1 Tournament Feedback Arc Set . . . . . . . . . . . . . . . . . . . . 52

6.1.1 Color and Conquer . . . . . . . . . . . . . . . . . . . . . . 526.1.2 Kernelization . . . . . . . . . . . . . . . . . . . . . . . . . 536.1.3 Probability of a Good Coloring . . . . . . . . . . . . . . . 546.1.4 Solving a Colored Instance . . . . . . . . . . . . . . . . . . 54

6.2 Universal Coloring Families . . . . . . . . . . . . . . . . . . . . . 57

7 Explicit Identification in W[1]-Hardness Reductions 617.1 Hardness for Treewidth-Parameterizations . . . . . . . . . . . . . 61

7.1.1 Equitable Coloring . . . . . . . . . . . . . . . . . . . . . . 627.1.2 Capacitated Dominating Set . . . . . . . . . . . . . . . . . 67

7.2 Hardness for Cliquewidth-Parameterizations . . . . . . . . . . . . 717.2.1 Graph Coloring — Chromatic Number . . . . . . . . . . . 717.2.2 Edge Dominating Set . . . . . . . . . . . . . . . . . . . . . 747.2.3 Hamiltonian Cycle . . . . . . . . . . . . . . . . . . . . . . 797.2.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

8 Kernelization Techniques 858.1 Meta-Theorems for Kernelization . . . . . . . . . . . . . . . . . . 85

8.1.1 Reduction Rules . . . . . . . . . . . . . . . . . . . . . . . 878.1.2 Decomposition Theorems . . . . . . . . . . . . . . . . . . . 948.1.3 Kernels . . . . . . . . . . . . . . . . . . . . . . . . . . . . 988.1.4 Implications of Our Results . . . . . . . . . . . . . . . . . 1018.1.5 Practical considerations: . . . . . . . . . . . . . . . . . . . 109

8.2 Turing Kernels . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1108.2.1 The Rooted k-Leaf Out-Branching Problem . . . . . . . . 1118.2.2 Bounding Entry and Exit Points . . . . . . . . . . . . . . . 1158.2.3 Bounding the Length of a Path . . . . . . . . . . . . . . . 118

9 Kernelization Lower Bounds 1259.1 Polynomial Parameter Transformations with Turing Kernels . . . 1259.2 Explicit Identification in Composition Algorithms . . . . . . . . . 128

9.2.1 Steiner Tree, Variants of Vertex Cover, and Bounded RankSet Cover . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

9.2.2 Unique Coverage . . . . . . . . . . . . . . . . . . . . . . . 1349.2.3 Bounded Rank Disjoint Sets . . . . . . . . . . . . . . . . . 1369.2.4 Domination and Transversals . . . . . . . . . . . . . . . . 1379.2.5 Numeric Problem: Small Subset Sum . . . . . . . . . . . . 140

10 Concluding Remarks and Open Problems 143

Chapter 1

Introduction

Parameterized Algorithms and Complexity is a natural way to cope with problemsthat are considered intractable according to the classical P versus NP dichotomy.In practice, large instances of NP-complete problems are solved exactly everyday. The reason for why this is possible is that problem instances that occurnaturally often have a hidden structure. The basic idea of Parameterized Algo-rithms and Complexity is to extract and harness the power of this structure. InParameterized Algorithms and Complexity every problem instance comes with arelevant secondary measurement k, called a parameter. The parameter measureshow difficult the instance is, the larger the parameter, the harder the instance.The hope is that at least for small values of the parameter, the problem instancescan be solved efficiently. This naturally leads to the definition of fixed param-eter tractability. We say that a problem is fixed parameter tractable (FPT) ifproblem instances of size n can be solved in f(k)nO(1) time for some function findependent of n.

Over the last two decades, Parameterized Algorithms and Complexity hasestablished itself as an important subfield of Theoretical Computer Science. Pa-rameterized Algorithms has its roots in the Graph Minors project of Robertsonand Seymour [122], but in fact, such algorithms have been made since the earlyseventees [54, 108]. In the late eightees several FPT algorithms were discov-ered [61, 59] but despite considerable efforts no fixed parameter tractable algo-rithm was found for problems like Independent Set and Dominating Set.In the early ninetees, Downey and Fellows [51, 50] developed a framework forshowing infeasibility of FPT algorithms for parameterized problems under cer-tain complexity theoretic assumptions, and showed that the existence of FPTalgorithms for Independent Set and Dominating Set is unlikely.

The possibility of showing lower as well as upper bounds sparked interest inthe field, and considerable progress has been made in developing methods to showtractability and intractability of parameterized problems. Notable methods thathave been developed include Bounded Search Trees, Kernelization, Color Coding

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and Iterative Compression. Over the last few years Kernelization has grown into asubfield of its own, due to a connection to polynomial time preprocessing and thetechniques that have been developed over the last year for showing kernelizationlower bounds [70, 20].

In this thesis we develop new methods for showing upper and lower bounds,both for Parameterized Algorithms and for Kernelization.

1.1 Overview of the Thesis

In Section 1.2 we give the necessary notation and definitions for the thesis. InPart I we give an overview of existing techniques in Parameterized Algorithmsand Complexity. This part of the thesis is organized as follows. First we covertechniques for algorithm design, then we review existing medthods for show-ing running time lower bounds up to certain complexity-theoretic assumptions.Then we turn our attention to kernelization or polynomial time preprocessing.In Chapter 4 we describe the most well-known techniques for proving kerneliza-tion results. In Chapter 5 we survey the recent methods developed for givingkernelization lower bounds.

In Part II we describe new methods in Parameterized Algorithms and Com-plexity that were developed as part of the author’s PhD research. The part isorganized similarly to Part I, that is, we first consider algorithmic upper andlower bounds, before turning our attention to upper and lower bounds for ker-nelization. In Chapter 6 we describe a new color-coding based scheme to givesubexponential time algorithms for problems on dense structures. In Chapter 7we illustrate how explicit identification makes it easier to construct hardness re-ductions, and use this kind of reductions to show a trade-off between expressivepower and computational tractability. In Chapter 8 we give two results. Our firstresult is a meta-theorem for kernelization of graph problems restricted to graphsembeddable into surfaces with constant genus. Our second result is a demonstra-tion that a relaxed notion of kernelization called Turing kernelization indeed ismore powerful than kernelization in the traditional sense. Finally, in Chapter 9we show that explicit identification is useful not only in hardness reductions, butalso to show kernelization lower bounds. Several such lower bounds are derivedby applying the proposed method.

Parts of this thesis have been published as conference articles and some partsare excerpts from articles currently under peer review. Below we give an overviewof the articles that were used as a basis for this work.

• Section 2.3 is based on the manuscript Simpler Parameterized Algorithmfor OCT, coauthored with S. Saurabh and S. Sikdar.

• Section 2.8 is based on an excerpt from the article Graph Layout problems

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Parameterized by Vertex Cover, coauthored with M. Fellows, N. Misra, F.A. Rosamond and S. Saurabh. Proceedings of ISAAC 2008, pages 294-305.

• Section 4.2 is based on an excerpt from the manuscript Hypergraph Coloringmeets Hypergraph Spanning Tree, coauthored with S. Saurabh.

• Chapter 6 is based on the manuscript Fast FAST, coauthored with N. Alonand S. Saurabh.

• Section 7.1.1 is based on an excerpt from the article On the Complexityof Some Colorful Problems Parameterized by Treewidth, coauthored withM. Fellows, F. V. Fomin, F. Rosamond, S. Saurabh, S. Szeider and C.Thomassen. In the proceedings of COCOA 2007, pages 366-377. (InvitedPaper.)

• Section 7.1.2 is based on an excerpt from the article Capacitated Dominationand Covering: A Parameterized Perspective, coauthored with M. Dom, Y.Villanger and S. Saurabh. Proceedings of IWPEC 2008, pages 78-90.

• Section 7.2 is based on the article Clique-width: On the Price of Generality,coauthored with F. V. Fomin, P. A. Golovach and S.Saurabh. Proceedingsof SODA 2009, pages 825-834.

• Section 8.1 is based on the manuscript (Meta)-Kernelization, coauthoredwith H. Bodlaender, F. V. Fomin, E. Penninkx, S. Saurabh and D. Thilikos.

• Sections 8.2 and 9.1 are based on the article Kernel(s) for Problems With noKernel: On Out-Trees With Many Leaves, coauthored with H. Fernau, F.V. Fomin, D. Raible, S. Saurabh and Y. Villanger. Proceedings of STACS2009, pages 421-432.

• Section 9.2 is based om the manuscript Incompressibility through Colorsand IDs, coauthored with M. Dom and S. Saurabh.

1.2 Notation and Definitions

Basic Notions We assume that the reader is familiar with basic notions likesets, functions, polynomials, relations, integers etc. In particular, for these no-tions we follow the setup of [116].

Problems A problem or language is a set L of strings over a finite alphabetΣ, that is L ⊆ (Σ)∗. A string s ∈ L is a yes instance of L and a string s /∈ Lis a no instance of L. A parameterized problem is a subset Π of (Σ)∗ × N, thatis, every instance of Π comes with a natural number called the parameter. In an

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optimization problem we are given as input an instance s, defining a set F (s) offeasible solutions for a problem specific function F . The objective is to find anelement x ∈ F (s) that minimizes or maximizes a certain problem spesific objectivefunction g(x). If the objective is to minimize (maximize) g(x) the problem isreferred to as a minimization (maximization) problem. The minimum (maximum)value of the objective function over all the feasible solutions for an instance s iscalled the optimal value of the objective function.

Function Growth We employ big-Oh notation which suppresses constant fac-tors and lower order terms. That is, for two functions f : N → N and g : N → N

we say f(n) = O(g(n)) if there are constants c0, c1 > 0 such that for all x ≥ c0,f(x) ≤ c1g(x). We say that a function f is singly exponential if f = O(2nc

) for

some constant c. If f = O(22nc

) for some constant c we say f is doubly exponen-tial. We will sometimes misuse notation and say that f(n) is at least O(g(n)),meaning that g(n) = O(f(n)).

Algorithms An algorithm for the problem L is an algorithm that for a strings determines whether s ∈ L. The running time of the algorithm is measured inthe number of steps the algorithm performs. We will assume a single processor,random-access machine as the underlying machine model throughout this thesis.In the random-access machine any simple operation (arithmetic, if-statements,memory-access etc.) takes unit length of time, and the word size is sufficientlylarge to hold numbers that are singly exponential in the size of the input. Ac-approximation algorithm for a minimization problem is an algorithm that for agiven instance finds a feasible solution x such that the value of the of the objectivefunction g(x) is at most c times the optimal value. Similarly a c-approximationalgorithm for a maximization problem is an algorithm that for a given instancefinds a feasible solution x such that the value of the of the objective functiong(x) is at least 1/c times the optimal value. An optimization problem that hasa c-approximation algorithm is said to be c-approximable. A polynomial timeapproximation scheme (PTAS) for an optimization problem is an algorithm thatfor any fixed ǫ > 0 computes a polynomial time (1 + ǫ)-approximation algorithmfor the problem.

Graphs A graph G is a set V (G) of vertices and a set E(G) of unorderedpairs of vertices called edges. A vertex u and a vertex v are adjacent if theedge uv ∈ E(G), and the vertices u and v are incident to the edge uv. Thevertices u and v are referred to as the endpoints of the edge uv. If vv /∈ E(G) forall v ∈ V (G) the graph G is called simple. Unless specifically stated otherwisethe graphs considered in this thesis are simple. The open neighbourhood or justneighbourhood of a vertex v in the graph G is the set NG(v) = u : uv ∈ E(G)

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and the closed neighbourhood of a vertex v is the set NG[v] = NG(v) ∪ v. Theneighbourhood of a vertex set S is NG(S) = (

⋃v∈S N(v)) \ S. For a vertex set

S ⊂ V (G), we define ∂G(S) as the set of vertices in S that have a neighbor inV \S. The degree of a vertex v is |NG(v)|. For v ∈ V (G), by EG(v) we mean theset of edges incident to v. When the graph is clear from the context we will omitthe subscripts.

A walk in a graph G is a sequence v1, . . . , vk of vertices such that vivi+1 ∈E(G). A walk that never contains the same vertex twice is called a path. A walkwhere the first and the last vertex is the same but the other vertices are uniqueis called a cycle. The length of a walk is k − 1. A subgraph of G is a graph G′

such that V (G′) ⊆ V (G) and E(G′) ⊆ E(G). The subgraph induced by a vertexset S is G[S] = (S, uv ∈ E(G) : u ∈ S ∧ v ∈ S). For an edge set S, byG \ S we denote G′ = (V (G), E(G) \ S) and for a vertex set S ′, G \ S we denoteG[V (G) \ S]. For a vertex x and edge e, G \ x and G \ e is defined as G \ xand G \ e respectively. Contracting an edge uv of G amounts to removingthe vertices u and v from the graph and adding a new vertex u′ and making u′

adjacent to N(u) ∪ N(v) \ u, v. If H can be obtained from G by repeatedlycontracting edges, we say that H is a contraction of G. If H is a subgraph of acontraction of G we say that H is a minor of G, and that H ≤M G. A graph Gis connected if there is a path between every pair of vertices. We will say thata vertex or edge set with a specific property is minimal (maximal) if no propersubset (superset) of the set has the property. A connected component of G is aninduced subgraph C of G such that V (C) is a maximal subset of V (G) such thatG[V (C)] is connected. A set S ⊆ V (G) is a separator if G \ S is disconnected.

A set S ⊆ V (G) of pairwise adjacent vertices is called a clique and a set of aset S ⊆ V (G) of pairwise non-adjacent vertices is called an independent set. Aset S ⊆ V (G) such that every edge has an endpoint in S is called a vertex coverof G. A set S such that every vertex in V (G) \ S has a neighbour in V is calleda dominating set of G.

Treewidth A tree decomposition of a graph G is a pair (X, T ) where T is a treewhose vertices we will call nodes and X = (Xi | i ∈ V (T )) is a collection ofsubsets of V (G) such that

1.⋃

i∈V (T )Xi = V (G),

2. for each edge vw ∈ E(G), there is an i ∈ V (T ) such that v, w ∈ Xi, and3. for each v ∈ V (G) the set of nodes i | v ∈ Xi forms a subtree of T .

The width of a tree decomposition (Xi | i ∈ V (T ), T ) equals maxi∈V (T )|Xi| −1. The treewidth of a graph G is the minimum width over all tree decompositionsof G. We use notation tw(G) to denote the treewidth of a graph G. A treedecomposition is called a nice tree decomposition if the following conditions aresatisfied: Every node of the tree T has at most two children; if a node t has two

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children t1 and t2, then Xt = Xt1 = Xt2 ; and if a node t has one child t1, theneither |Xt| = |Xt1 | + 1 and Xt1 ⊂ Xt or |Xt| = |Xt1 | − 1 and Xt ⊂ Xt1 . It ispossible to transform a given tree decomposition into a nice tree decompositionin time O(|V | + |E|) [18].

Clique-width Let G be a graph, and k be a positive integer. A k-graph is agraph whose vertices are labeled by integers from 1, 2, . . . , k. We call the k-graph consisting of exactly one vertex labeled by some integer from 1, 2, . . . , kan initial k-graph. The cliquewidth cwd(G) is the smallest integer k such that ak-graph isomorphic to G can be constructed by means of repeated application ofthe following four operations:

1. Introduce, construction of an initial k-graph labeled by i, denoted by i(v).

2. Disjoint union, denoted by ⊕.

3. Relabel, changing all labels i to j, denoted by ρi→j.

4. Join, connecting all vertices labeled by i with all vertices labeled by j byedges, denoted by ηi,j.

An expression tree of a graph G is a rooted tree T of the following form:

• The nodes of T are of four types i, ⊕, η and ρ.

• Introduce nodes i(v) are leaves of T , corresponding to initial k-graphs withvertices v, which are labeled i.

• A union node ⊕ stands for a disjoint union of graphs associated with itschildren.

• A relabel node ρi→j has one child and is associated with the k-graph, whichis the result of relabeling operation for the graph corresponding to the child.

• A join node ηi,j has one child and is associated with the k-graph, which isthe result of join operation for the graph corresponding to the child.

• The graph G is isomorphic to the graph associated with the root of T (withall labels removed).

The w idth of the tree T is the number of different labels appearing in T . If agraph G has cwd(G) ≤ k then it is possible to construct a rooted expressiontree T with width k of G. A well-known fact is that if the treewidth of a graphis bounded then its cliquewidth also is bounded. On the other hand, completegraphs have clique-width 2 and unbounded treewidth.

Theorem 1.2.1 ([34]) If graph G has treewidth at most t, then cwd(G) is atmost k = 3 · 2t−1. Moreover, an expression tree for G of width at most k canbe constructed in FPT time (with treewidth being the parameter) from the treedecomposition of G.

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The second claim in Theorem 1.2.1 is not given explicitly in [34]. Howeverit can be shown since the upper bound proof in [34] is constructive (see also[37, 57]). Note that if a graph has bounded treewidth then the correspondingtree decomposition can be constructed in linear time [18].

Graphs on Surfaces We refer to the textbook “Graphs on Surfaces” by Moharand Thomassen [110] for an introduction to the topological graph theory usedhere. Let Gg be the class of all graphs that can be embedded into a surface Σ ofEuler-genus at most g. We say that a graph G is Σ-embedded if it is accompaniedwith an embedding of the graph into Σ. The radial distance between x and yis defined to be one less than the minimum length of a sequence starting fromx and ending at y such that vertices and faces alternate in the sequence. Givenan Σ-embedded graph G = (V,E) and a set S ⊆ V , we denote by Rr

G(S) andBr

G(S) the set of all vertices that are in radial distance at most r and distanceat most r from some vertex in S respectively. Notice that for every set S ⊆ Vand every r ≥ 0, it holds that Br

G(S) ⊆ R2r+1G (S) for any embedding of G into a

surface Σ. An alternative way of viewing radial distance is to consider the radialgraph, RG: an embedded multigraph whose vertices are the vertices and the facesof G (each face f of G is represented by a point vf in it). An edge between avertex v and a vertex vf is drawn if and only if v is incident to f . Thus RG is abipartite multigraph, embedded in the same surface as G. Hence, if G ∈ Gg thenRG ∈ Gg. Also the radial distance between a pair of vertices in G corresponds tothe normal distance in RG.

t-Boundaried Graphs We define the notion of t-boundaried graphs and variousoperations on them.

Definition 1.2.2 [t-Boundaried Graphs] A t-boundaried graph is a graph G =(V,E) with t distinguished vertices, uniquely labelled from 1 to t. The set ∂(G)of labelled vertices is called the boundary of G. The vertices in ∂(G) are referredto as boundary vertices or terminals.

For a graph G = (V,E) and a vertex set S ⊆ V , we will sometimes considerthe graph G[S] as the |∂(S)|-boundaried graph with ∂(S) being the boundary.

Definition 1.2.3 [Gluing by ⊕] Let G1 and G2 be two t-boundaried graphs. Wedenote by G1⊕G2 the t-boundaried graph obtained by taking the disjoint union ofG1 and G2 and identifying each vertex of ∂(G1) with the vertex of ∂(G2) with thesame label; that is, we glue them together on the boundaries. In G1 ⊕G2 there isan edge between two labelled vertices if there is an edge between them in G1 or inG2.

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Definition 1.2.4 [Legality] Let G be a graph class, G1 and G2 be two t-boundariedgraphs, and G1, G2 ∈ G. We say that G1 ⊕ G2 is legal with respect to G if theunified graph G1 ⊕G2 ∈ G. If the class G is clear from the context we do not saywith respect to which graph class the operation is legal.

Definition 1.2.5 [Replacement] Let G = (V,E) be a graph containing an ex-tended r-protrusion X. Let X ′ be the restricted protrusion of X and let G1 be anr-boundaried graph. The act of replacing X ′ with G1 corresponds to changing Ginto G[V \X ′] ⊕ G1. Replacing G[X] with G1 corresponds to replacing X ′ withG1.

Finite Integer Index

Definition 1.2.6 For a parameterized problem, Π on a graph class G and twot-boundaried graphs G1 and G2, we say that G1 ≡Π G2 if there exists a constantc such that for all t-boundaried graphs G3 and for all k,

• G1 ⊕G3 is legal if and only if G2 ⊕G3 is legal.

• (G1 ⊕G3, k) ∈ Π if and only if (G2 ⊕G3, k + c) ∈ Π.

Definition 1.2.7 [Finite Integer Index] Π has finite integer index in G if forevery t there exists a finite set S of t-boundaried graphs such that S ⊆ G and forany t-boundaried graph G1 there exists a G2 ∈ S such that G2 ≡Π G1. Such a setS is called a set of representatives for (Π, t)

Note that for every t, the relation ≡Π on t-boundaried graphs is an equivalencerelation. A problem Π is finite integer index, if and only if for every t, ≡Π is offinite index, that is, has a finite number of equivalence classes. The term finiteinteger index first appeared in the work by Bodlaender and van Antwerpen-deFluiter [25, 41] and is similar to the notion of finite state [2, 27, 38].

Digraphs A digraph or directed graph D is a set V (D) of vertices and a setA(D) of ordered pairs of vertices called arcs. Given a subset V ′ ⊆ V (D) of adigraph D, by D[V ′] we mean the digraph induced on V ′. A vertex y of D is anin-neighbor (out-neighbor) of a vertex x if yx ∈ A (xy ∈ A). The in-degree (out-degree) of a vertex x is the number of its in-neighbors (out-neighbors) in D. Apath in a digraph is a sequence P = p1p2 . . . pℓ of vertices such that pipi+1 ∈ A(D)for every i. Let P = p1p2 . . . pℓ be a given path. Then by P [pipj] we denote asubpath of P starting at vertex pi and ending at vertex pj . A cycle in D isa sequence C = c1c2 . . . cℓ of vertices such that cici+1 ∈ A(D) for every i andcℓc1 ∈ A(D).

A digraph D is acyclic if D does not contain any cycles. It is well knownthat a digraph D is acyclic if and only if the vertices of D can be ordered into

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v1 . . . vn such that for every arc vivj ∈ A(D) we have i < j. Such an ordering iscalled a topological ordering of D. A feedback arc set of a digraph D is an arcset S such that D \ S is acyclic. An out-tree is an acyclic digraph where everyvertex except one has indegree 1 and one vertex, called the root with indegree0. An out-branching in a digraph D is an out-tree in D containing all vertices ofD. For a given vertex q ∈ V (D), by q-out-branching (or q-out-tree) we denote anout-branching (out-tree) of D rooted at vertex q.

We say that the removal of an arc uv (or a vertex set S) disconnects a vertexw from the root r if every path from r to w in D contains arc uv (or one of thevertices in S). An arc uv is contracted as follows: add a new vertex u′, and foreach arc wv or wu add the arc wu′, and for an arc vw or uw add the arc u′w,remove all arcs incident to u and v and the vertices u and v.

Let T be an out-tree of a digraph D. We say that u is a parent of v and v isa child of u if uv ∈ A(T ). We say that u is an ancestor of v if there is a directedpath from u to v in T . An arc uv in A(D) \A(T ) is called a forward arc if u is anancestor of v, a backward arc if v is an ancestor of u and a cross arc, otherwise.

A tournament is a digraph T where every pair of vertices is connected byexactly one arc. An arc weighted tournament is a tournament which comes witha weight function w : A → R. For an arc weighted tournament we define theweight function w∗ : V × V → R such that w∗(u, v) = w(uv) if uv ∈ A and 0otherwise. Given a directed graph D = (V,A) and a set F of arcs in A defineDF to be the directed graph obtained from D by reversing all arcs of F . Thefollowing is a useful characterization of minimal feedback arc sets in directedgraphs.

Proposition 1.2.8 ([119]) Let D = (V,A) be a directed graph and F be a subsetof A. Then F is a minimal feedback arc set of D if and only if F is a minimalset of arcs such that DF is a directed acyclic graph.

Given a minimal feedback arc set F of a tournament T , the ordering σ cor-responding to F is the unique topological ordering of TF. Conversely, givenan ordering σ of the vertices of T , the feedback arc set F corresponding to σ isthe set of arcs whose endpoint appears before their startpoint in σ. The cost ofan arc set F is

∑e∈F w(e) and the cost of a vertex ordering σ is the cost of the

feedback arc set corresponding to σ.

Types of Logic The syntax of Monadic Second Order Logic (MSO2) of graphsincludes the logical connectives ∨, ∧, ¬, ⇔, ⇒, variables for vertices, edges, setof vertices and sets of edges, the quantifiers ∀, ∃ that can be applied to thesevariables, and the following five binary relations: (1) u ∈ U where u is a vertexvariable and U is a vertex set variable; (2) d ∈ D where d is an edge variable andD is an edge set variable; (3) inc(d, u), where d is an edge variable, u is a vertex

10

variable, and the interpretation is that the edge d is incident on the vertex u; (4)adj(u, v), where u and v are vertex variables u, and the interpretation is that uand v are adjacent; (5) equality of variables representing vertices, edges, set ofvertices and set of edges. If we do not have variables for sets of edges, this typeof logic is called MSO1 logic. If we only allow variables for vertices and edges, i.ethere are no set variables we get First Order (FO) logic. An extention of MSO2

called counting monadic second-order logic or CMSO is obtained from MSO2 byalso allowing atomic formulas for testing whether the cardinality of a set is equalto n modulo p, where n and p are integers such that 0 ≤ n < p and p ≥ 2.Essentially, CMSO is just MSO2 equipped with the following atomic formula: IfU denotes a set X, then cardn,p(U) = true if and only if |X| is n mod p. Itis known that every set F of graphs of bounded treewidth is CMSO-definable ifand only if F is finite state [101].

Vectors For a pair of integer row vectors p = [p1, . . . , pt], q = [q1, . . . , qt] wesay that p ≤ q if pi ≤ qi for all i. The transpose of a row vector p is denotedby p†. The t-sized vector e is [1, 1, . . . , 1], 0 is [0, 0, . . . , 0] and ei is the t-sizedvector with all entries 0 except for the i’th which is 1. Let O(

√k) denote, as

usual, any function which is O(√k(log k)O(1)). For any positive integer m put

[m] = 1, 2, . . . , m.

Part I

Overview of the Field

11

Chapter 2

Algorithmic Techniques

2.1 Bounded Search Trees

Bounded Search Trees is a simple yet powerful technique that has found manyapplications in Parameterized Algorithm Design [116]. The basic idea is to reducea given instance (I, k) to f(k) independent instances (I1, k1), . . . (If(k), kf(k)) inpolynomial time such that

• For every 1 ≤ i ≤ f(k) we have ki < k and |Ii| ≤ |I| · g(k) for a function g.

• Instances (I, 0) are polynomial time solvable.

• The functions f and g are non-decreasing and depend only on k.

While the definition above might seem somewhat involved, most algorithmsthat apply the Bounded Search Tree technique do so with f(k) ≤ ko(1) andg(k) = 1. One might even argue that usually f(k) is a constant independent ofk. Let (I, k) be the considered instance. If k > 0 the algorithm performs thereduction to (I1, k1), . . . (If(k), kf(k)) in polynomial time and makes recursive callsto (Ii, ki) for every i between 1 and f(k). Let c be a constant such that reducingan instance to the f(k) smaller ones and solving instances (I, 0) with |I| ≤ n bothtake O(nc) time, and let T (n, k) be an upper bound on the running time of thealgorithm on an instance (I, k). Then T (n, k) ≤ f(k) · T (g(k)n, k − 1) + O(nc).Expanding this recurrence k times yields T (n, k) = O((f(k)g(k)c)k · nc)). Asan example we demonstrate how to apply this technique to get a simple FPTalgorithm for the Vertex Cover problem where we are given a gragh G and aninteger k and asked whether there is a subset S of V (G) of size at most k suchthat every edge in E(G) has at least one endpoint in S.

Theorem 2.1.1 ([108, 52]) There is an O(2kn)-time algorithm for VertexCover.

13

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Proof. First observe that a graph G has a vertex cover of size 0 if and only if Ghas no edges. Consider now an instance (G, k) of Vertex Cover with k > 0and an edge uv ∈ E(G). Any vertex cover S of G contains either u or v sinceS must contain at least one endpoint of uv. A set S that contains a vertex xis a vertex cover of G if and only if S \ x is a vertex cover of G \ x since thevertex x covers all edges incident to it. Hence G has a vertex cover of size atmost k if and only if G \u or G \ v has a vertex cover of size at most k− 1. Thusthere is a polynomial time reduction from the instance (G, k) to the two instances(G \ u, k − 1), (G \ v, k − 1). The functions f and g in the discussion above aref(k) = 2 and g(k) = 1 and hence this algorithm runs in time O(2k · n +m).

The above algorithm first appeared already in 1984 in a monograph by Mel-horn [108]. However, this went unnoticed by the FPT community. In the “Pa-rameterized Complexity” monograph of Downey and Fellows [52], which waspublished in 1999, the history of the development of algorithms for the Ver-tex Cover problem is described. In 1987, Fellows and Langston showed that asa consequence of deep theorems in the Graph Minors project of Robertson andSeymour, Vertex Cover can be solved in O(n3) time for every fixed value ofk [61]. Later the same year, Johnson showed that the Vertex Cover prob-lem can be solved in time O(n2) for every fixed value of k. Neither Fellows andLangston nor Johnson give explicitely the dependence of the running time on k,and this dependence is at least doubly exponential in k. In 1988, Fellows [59]independently rediscovered the O(2kn + m) algorithm from Theorem 2.1.1. In1989, Buss [52] described an algorithm with running time O(kn+ 2kk2k+2). Oneshould notice that in this algorithm, the dependence on k is worse than in thealgorithm of Theorem 2.1.1, but that this dependence is separated completelyfrom the dependence on n, that is, this term is additive instead of multiplicative.In 1992, combining the ideas of Buss and the algorithm in [59], Balasubramanianet al. [13] gave an algorithm with running time O(kn + 2kk2). In 1996, Bala-subramanian et al. [13] gave an algorithm with running time O(kn + (4

3)kk2).

Independently of this development, in 1993 Papadimitriou and Yannakakis [118]gave an algorithm with running time O(3kn), based on matching techniques. Ina series of papers, the ideas in [13] were refined, and to this date the best knownalgorithm for Vertex Cover is due to Chen et al. [30] and takes O(1.274k +kn)time. In the next section we give a brief introduction to some ideas that can beused to improve the running time of bounded search tree algorithms.

2.2 Reduction Rules and Branching Recurrences

A reduction rule is a rule or a polynomial time algorithm that transforms aninstance (I, k) of a parameterized problem to an “equivalent and simpler” instance(I ′, k′). Here equivalent means that (I, k) is a yes instance if and only if (I ′, k′)

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is and the precise definition of what “simpler” means varies from problem toproblem. It could mean that |I ′| < |I|, it could mean that k′ < k or it couldmean that the instance I ′ contains fewer occurances of a particular substructure.We will say that a reduction rule is correct if the instances (I, k) and (I ′, k′)indeed are equivalent.

A good thing about reduction rules is that as long as one is only interestedin whether (I, k) is a yes or no instance, there is absoutely no reason for notperforming the reduction rule. Even better, quite often we can give performanceguarantees on our reduction rules and show that an instance (I, k) to which ourreduction rules can not be applied must satisfy |I| ≤ f(k) for some function f . Inthis case we say that the problem admits an f(k) − kernel. The study of kernelshas developed to become a flourishing subfield of Parameterized Algorithms andComplexity. A more thorough introduction to kernelization is deferred to Chapter4. Reduction rules can often be useful independently of whether they give a kernelfor the problem. In particular combining reduction rules with the Bounded SearchTree technique often gives improved running time bounds over a plain BoundedSearch Tree algorithm. We illustrate this by describing an algorithm for VertexCover with running time better than the algorithm from Theorem 2.1.1.

Theorem 2.2.1 (Folklore) There is an O(1.6181kn2) time algorithm for VertexCover.

Proof. We apply the following two rules: If G has a vertex u with degree 0then let (G′, k′) = (G \ u, k). If G has a vertex u with degree 1, let (G′, k′) =(G \N [u], k − 1). Correctness of the first rule is obvious, and the second rule iscorrect because any vertex cover containing u could be swapped with a vertexcover containing N(u) instead. If neither rule can be applied, all vertices havedegree at least 2. Pick a vertex u. Any vertex cover of G must contain either u orN(u). Thus (G, k) is a yes instance to Vertex Cover if and only if (G\u, k−1)or (G \N(u), k − |N(u)|) is. Let T (n, k) is an upper bound on the running timeof our algorithm. Since |N(u)| ≥ 2 we have T (n, k) ≤ T (n, k− 1)+T (n, k− 2)+O(n2). Thus T (n, k) ≤ O(λkn2) where λ is the largest root of the characteristicpolynomial λk−λk−1−λk−2 of the recurrence. The largest root of this polynomialis the golden ratio (1 +

√5)/2 ≤ 1.6181 so T (n, k) ≤ O(1.6181kn2) completing

the proof.

A trait of branching algorithms is that the running time bound often can beimproved by a refined case analysis. The positive aspect of this is that quite goodrunning time bounds are achievable. A drawback is that considering more andmore cases in the algorithm reduces clarity and makes the constant hidden inthe big-Oh notation larger. Simulations have shown that in practice, the moreadvanced reduction and branching rules often slow the algorithm down [128].

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2.3 Iterative Compression

Iterative Compression is a tool that has recently been used successfully to giveFPT algorithms for a number of problems. This technique was first introducedby Reed, Smith and Vetta in order to solve the Odd Cycle Transversalproblem [120]. In this problem we are given a graph G together with an integerk. The objective is to find a set S of at most k vertices whose deletion makes thegraph bipartite, and a set S such that G \ S is bipartite is called an odd cycletransversal of G. The method of Iterative Compression was used in obtainingfaster fixed parameter tractable (FPT) algorithms for Feedback Vertex Set,Edge Bipartization, Chordal Deletion and Cluster Vertex Deletionon undirected graphs [42, 77, 107, 85]. The technique was also used by Chen etal. [31] to show that the Directed Feedback Vertex Set problem is FPT,resolving a long standing open problem in Parameterized Complexity. In thissection we present a reinterpretation of the algorithm given by Reed, Smith andVetta for Odd Cycle Transversal. In particular, we give a new proof of thefollowing theorem.

Theorem 2.3.1 There is an algorithm that given a graph G = (V,E) and integerk decides whether G has an odd cycle transversal of size at most k in time O(3k ·k · |E| · |V |).

Proof. The idea is to reduce the problem in question to a modified version,where we are also given as input a solution that is almost good enough, but notquite. For the case of Odd Cycle Transversal, we are given an odd cycletransversal S ′ of G of size k+1. We call this problem the compression version ofOdd Cycle Transversal. The crux of the Iterative Compression method isthat often the compression version of a problem is easier to solve than the originalone.

Suppose we could solve the compression version of the problem in O(f(k)nc)time. We show how to solve the original problem in O(f(k)nc+1) time. Order thevertices of V (G) into v1v2 . . . vn and define Vi = v1 . . . vi for every i. Notice thatif G has an odd cycle transversal S of size k then S∩Vi is an odd cycle transversalof G[Vi] for every i ≤ n. Furthermore, if S is an odd cycle transversal of G[Vi]then S ∪ vi+1 is an odd cycle transversal of G[Vi+1]. Finally, Vk is an odd cycletransversal of size k ofG[Vk]. These three facts together with the f(k)nc algorithmfor the compression version of Odd Cycle Transversal give an f(k)nc+1 timealgorithm for Odd Cycle Transversal as follows. Call the algorithm forthe compression version with input (G[Vk+1], Vk+1, k). The algorithm will eitherreport that (G[Vk+1, k]) has no odd cycle transversal of size k or return such anodd cycle transversal, call it Sk+1. In the first case G has no odd cycle transversalof size k. In the second, call the algorithm for the compression version with input(G[Vk+2], Sk+1 ∪ vk+2, k). Again we either receive a “no” answer or a odd cycle

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transversal Sk+2 of G[Vk+2] of size k and again, if the answer is negative then Ghas no k-sized odd cycle transversal. Otherwise we call the compression algorithmwith input (G, Sk+2 ∪ vk+3, k) and keep going on in a similar manner. If wereceive a negative answer at some step we answer that G has no k-sized oddcycle transversal. If we do not receive a negative answer at any step, then aftern− k calls to the compression algorithm we have a k-sized odd cycle transversalof G[Vn] = G. Thus we have resolved the input instance in time O(f(k)nc+1).We refer to [116] for a more thorough introduction to Iterative Compression.

We now show how to solve the compression version of Odd Cycle Transver-sal in time O(3k · k · |E|). From the discussion in the previous paragraph it willfollow that Odd Cycle Transversal can be solved in time O(3k · k · |E| · |V |).For two vertex subsets V1 and V2 of V (G) a walk from V1 to V2 is a walk with oneendpoint in V1 and the other in V2, or a single vertex in V1 ∩ V2. The following isa simple fact about bipartite graphs.

Fact 2.3.2 Let G = (V1⊎V2, E) be a bipartite graph with vertex bipartition V1⊎V2.Then

1. For i ∈ 1, 2, no walk from Vi to Vi has odd length.

2. No walk from V1 to V2 has even length.

Recall that we are given a graph G and an odd cycle transversal S ′ of G ofsize k + 1 and we have to decide whether G has an odd cycle transversal of sizeat most k. If such an odd cycle transversal S exists then there exists a partitionof S ′ into L⊎R⊎T , where T = S ′∩S and L and R are subsets of the left and rightbipartitions of the resulting graph. The algorithm iterates over all 3k partitionsof S into L⊎R⊎T . For each partition we run an algorithm that takes as input apartition of S ′ into L ⊎R ⊎ T , runs in O(k · |E|) time and decides whether thereexists a set of vertices T ′ of size at most k − |T | in G \ S ′ such that G \ (T ∪ T ′)is bipartite with bipartitions VL and VR such that L ⊆ VL and R ⊆ VR. In theremainder of this section we give such an algorithm. This algorithm together withthe outer loop over all partitions of S ′ yields the O(3k · k · |E|) time algorithm forthe compression step.

Before proceeding we do a simple “sanity check”. If there is an edge in G[L]or G[R] it is clear that X can not exist since then either VL or VR can not be anindependent set. Hence if there is an edge in G[L] or G[R] we can immediatelyskip to the next partition of S ′. Now, since G \ S ′ is bipartite, let A ⊎ B be abipartition of G\S ′. Let Al and Bl be the neighbors of L in A and B respectively.Similarly let Ar and Br be the neighbours of R in A and B respectively.

Claim 2.3.3 Let (G, S ′, k) be an instance of the compression version of OddCycle Transversal and let S ′ = L ⊎ R ⊎ T . If X ⊆ (V (G) \ S ′) is a set of

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vertices such that G \ (T ∪X) is bipartite with bipartitions VL and VR such thatL ⊆ VL and R ⊆ VR, then in G \ (S ′ ∪X), there are no paths between Al and Bl;Bl and Br; Br and Ar; and, Ar and Al.

Proof. Any path from Al to Bl in G \ (S ′ ∪ X) has odd length and can beextended to a walk from L to L of odd length in G′ \ (T ∪ X), contradictingFact 2.3.2. A symmetric argument shows that there are no paths between Br

and Ar in G \ (S ′ ∪X). Any path from Bl to Br in G \ (S ′ ∪X) must be of evenlength and can be extended to a walk in G \ (T ∪X) from L to R of even length,again contradicting Fact 2.3.2. A symmetric argument yields that there are nopaths between Ar and Al.

Claim 2.3.4 Let (G, S ′, k) be an instance of the compression version of OddCycle Transversal and let S = L ⊎ R ⊎ T such that G[L] and G[R] areindependent sets. Let X be a set of vertices in V (G)\S ′ such that in G\(S ′∪X),there are no paths between Al and Bl; Bl and Br; Br and Ar; and, Ar and Al.Then G \ (T ∪X) is bipartite with bipartitions VL and VR such that L ⊆ VL andR ⊆ VR.

Proof. Notice that every path from a vertex in L to another vertex in L withinner vertices only in V (G)\(S ′∪X) must have even length. Similarly every pathfrom a vertex inR to another vertex inR with inner vertices only in V (G)\(S ′∪X)must have even length and every path from a vertex in L to a vertex in R withinner vertices only in V (G) \ (S ′ ∪ X) must have odd length. Since G[L] andG[R] are independent sets it follows that if G \ (T ∪ X) is bipartite then it hasbipartitions VL and VR such that L ⊆ VL and R ⊆ VR. We now prove thatG \ (T ∪X) is bipartite.

Consider a cycle in G \ (T ∪X). If C does not contain any vertices of (L∪R)then |E(C)| is even since G \ S ′ is bipartite. Let v1, v2, . . . vt be the vertices of(L ∪ R) ∩ C in their order of appearance along C. Let v0 = vt, then we havethat |E(C)| =

∑t−1i=0 dc(vi, vi+1). But then E(C) must be even since the number

if indices i such that vi ∈ L and vi+1 ∈ R is equal to the number of indices j suchthat vj ∈ R and vj+1 ∈ L. This concludes the proof.

To check whether G \T has an odd cycle transversal X such that G \ (T ∪X)is bipartite with bipartitions VL and VR such that L ⊆ VL and R ⊆ VR weproceed as follows. Construct an auxiliary graph G from G \ S ′ by introducingtwo special vertices s, t and connecting s to each vertex in Al ∪Br and t to eachvertex in Ar ∪ Bl. The Claims 2.3.3 and 2.3.4 show that it is sufficient to checkwhether there is an st-separator in G of size at most k − |T ∪ T ′|. This can bedone using max flow in time O(k ·|E|). This discussion together with Claims 2.3.3and 2.3.4 complete the proof of Theorem 2.3.1

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2.4 Dynamic Programming and Courcelle’s The-

orem

The Dynamic Programming (DP) technique has found many applications in thedesign of algorithms, Parameterized Algorithms being no exception. When de-signing Dynamic Programming based parameterized algorithms, the key is to tryto keep the size of the DP table and the time needed to compute each cell in thetable down to f(k)nO(1).

A well-known example of Dynamic Prorgramming in Parameterized Algo-rithms is the Dreyfus-Wagner algorithm for the Steiner Tree problem [54]. Inthe Steiner Tree problem we are given a connected graph G together with asubset X of V (G). The set X is called a set of terminals and |X| = k. Theobjective is to find a subtree T of G containing all terminals, minimizing |V (T )|,the number of vertices in T . The Dreyfus-Wagner algorithm solves this problemin time O(3knO(1)). Interestingly, if you parameterize Steiner Tree problemby the maximum number of non-terminal vertices allowed in the solution, theproblem becomes intractable. This is discussed in detail in Section 3.3.

Parameterized problems where the parameter is the treewidth of the inputgraph G are often tackled by doing dynamic programming over the tree de-composition of G. In particular it has been shown that Independent Set,Vertex Cover and Dominating Set in graphs of treewidth at most k canbe solved in time O(2kkO(1)n), O(2kkO(1)n) and O(4kn) respectively [116]. Manyother problems admit fast dynamic programming algorithms in graphs of boundedtreewidth.

A very useful tool for showing that a problem is FPT parameterized bytreewidth is the celebrated Courcelle’s theorem, which states that every problemexpressible in Monadic Second Order Logic is fixed parameter tractable parame-terized by the treewidth of the input graph. For a graph predicate φ expressedin MSO2 let |φ| be the length of the MSO2 expression for φ.

Theorem 2.4.1 (Courcelle’s Theorem [35]) There is a function f : N×N →N and an algorithm that given a graph G together with a tree-decomposition of Gof width t and a MSO2 predicate φ decides whether φ(G) holds in time f(|φ|, t)n.

To apply Courcelle’s theorem on a specific problem we need to show that theproblem is expressible in monadic second order logic. For example consider theIndependent Set problem. Here we are given as input a graph G, a tree-decomposition ofG of width t and an integer k. The objective is to decide whetherG has an independent set of size at least k. We formulate the Independent Setproblem in MSO2. That is, a graph G has an independent set of size k if and

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only if the following predicate holds:

φ(G) = ∃v1, v2, . . . , vk ∈ V (G) : v1 6= v2 ∧ ¬adj(v1, v2), v1 6= v3 ∧ ¬adj(v1, v2),

. . . v2 6= v3 ∧ ¬adj(v2, v3), . . . vk−1 6= vk ∧ ¬adj(vk−1, vk)

Observe that the length of the predicate φ depends only on k, and not on the sizeof the graph G. Hence, by Theorem 2.4.1, there is an algorithm that given a graphG together with a tree-decomposition of G of width at most t and an integer kdecides whether G has an independent set of size at least k in time f(k, t)n forsome function f . In fact, it is not really necessary that the tree-decomposition ofG is given. Due to a result of Bodlaender [18], a tree-decomposition of width t ofa graph G of treewidth t can be computed in time f(t)n for some function f .

Theorem 2.4.2 (Bodlaender’s Theorem [18]) There is a function f : N →N and an f(t)n time algorithm that given a graph G and integer t decides whetherG has treewidth at most t, and if so, constructs a tree-decomposition of width atmost t.

Hence, combining Theorems 2.4.1 and 2.4.2 yields that there is a function f :N × N → N and an algorithm that given a graph G of treewidth t and a MSO2

predicate φ decides whether φ(G) holds in time f(|φ|, t)n. For example, by thediscussion in the previous paragraph there is an algorithm that given a graph Gof treewidth t and an integer k decides whether G has an independent set of sizeat most k in time f(t, k)n for some function f .

Theorem 2.4.1 has been generalized even further. For instance it has beenshown that MSO2-optimization problems are fixed parameter tractable parame-terized by the treewidth of the input graph. In a MSO2-minimization problemyou are given a graph G and a predicate φ in MSO2 that describes a propertyof a vertex (edge) set in a graph. The objective is to find a vertex (edge) set Sof minimum size such that φ(G, S) holds. In a MSO2-maximization problem theobjective is to find a set S of maximum size such that φ(G, S) holds.

Theorem 2.4.3 ([27, 12]) There is a function f : N × N → N and an algo-rithm that given a graph G of treewidth t and a MSO2 predicate φ finds a largest(smallest) set S such that φ(G, S) holds in time f(|φ|, t)n.

Turning our attention back to the Independent Set problem, we can observethat by applying Theorem 2.4.3 we can obtain a stronger result than from The-orem 2.4.1. In particular, the Independent Set can be expressed as a MSO2-maximization problem as follows:

max |S| s.t:

φ(G, S) = ∀v1, v2v1 = v2 ∨ ¬adj(v1, v2) holds

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Hence, by Theorem 2.4.3 there is an algorithm that given a graph G of treewidth tas input can find a maximum size independent set in time f(t)n for some functionf .

Another way to generalize Theorem 2.4.1 is to consider larger classes of graphsthan graphs of bounded treewidth. In particular, if we restrict the predicate φ toMSO1 logic, Theorem 2.4.3 can be extended to graphs of bounded cliquewidth.

Theorem 2.4.4 ([36]) There is a function f : N×N → N and an algorithm thatgiven a graph G and a clique-expression of G of width t and a MSO1 predicateφ finds a largest (smallest) set S such that φ(G, S) holds in time f(|φ|, t)n.

There are natural problems expressible in MSO2 but not in MSO1. Examplesinclude Hamiltonian Cycle and Max Cut. In Section 7.2 we show thatup to certain complexity-theoretic assumptions, one can not hope to be able togeneralize Theorem 2.4.4 to also handle all problems in MSO2.

2.5 Well Quasi Ordering

The set of natural numbers is well-ordered since every two natural numbers arecomparable, that is, for any two numbers a and b such that a 6= b, either a < bor b < a. An unusual way to describe a well-ordered set is to say that it containsno anti-chain of length at least 2, where an anti-chain is a sequence a1, a2, . . . ak

such that for every i 6= j we have ai 6= aj and neither ai < aj nor ai > aj . Wecan relax the notion of well-ordering, and say that a set S is well-quasi-orderedunder a relation < if every anti-chain in S is finite.

Robertson and Seymour, proved in their graph minors project that the setof graphs is well-quasi-ordered under the minor relation, thereby proving theGraph Minor Theorem and resolving Wagner’s Conjecture [135]. On the waythey defined many other interesting and useful concepts and showed results thathave been very useful for Parameterized Algorithms and Complexity. The conceptof treewidth was for instance introduced as a part of the graph minors project, andthe Bidimensionality theory discussed in Section 2.6 relies heavily on the graphminors project. In this section, however, we only discuss the direct implicationsof the Graph Minor Theorem to Parameterized Algorithms and Complexity.

Proposition 2.5.1 (Graph Minor Theorem [122]) The set of graphs is well-quasi-ordered under the minor relation.

Consider a graph class G that is closed under taking minors. That is, if G ∈ Gand H ≤M G then H ∈ G as well. Consider the set of graphs F consisting ofall graphs not in G such that all their minors is in G. We call this set the setof forbidden minors of G. Notice that every graph G that is not in G must have

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some minor in F . Notice also that by definition, F is an antichain, and that bythe Graph Minor Theorem F is finite. Hence, to check whether G belongs to Git is sufficient to check for every H ∈ F whether G contains H as a minor. Tothis end, the following theorem, also from the graph minors project, is useful.

Proposition 2.5.2 ([121]) For every fixed graph H there is an O(n3) time al-gorithm to check for an input graph G whether H ≤M G.

Combining the Graph Minors Theorem with Proposition 2.5.2 yields thatevery minor closed graph class G can be recognized in O(n3) time, where theconstant hidden in the big-Oh notation depends on G. To put this in termsof Parameterized Algorithms - the problem of deciding whether G belongs to aminor closed class G is FPT parameterized by G.

Corollary 2.5.3 ([61]) Vertex Cover is fixed parameter tractable.

Proof. We prove that if a graph G has a vertex cover C of size at most k thenso do all the minors of G. For any edge e ∈ E(G), C is a vertex cover of G \ e.Similarly for any v ∈ V (G), C \v is a vertex cover of G\v. Finally, for an edgeuv ∈ E(G) let Guv be the graph obtained by contracting the edge uv and let u′

be the vertex obtained from u and v during the contraction. Then, every edge inE(G) with u or v being one of its endpoints will have u′ as its endpoint in Guv.Hence, if neither u nor v are in C then C is a vertex cover of Guv. If u ∈ C orv ∈ C then (C \ u, v) ∪ u′ is a vertex cover of size at most k of Guv. Hence,for a fixed integer k ≥ 0 the class Ck of graphs that have a vertex cover of size atmost k is closed under taking minors.

By Theorem 2.5.1 Ck has a finite set Fk of forbidden minors. By Proposition2.5.2, for each H ∈ Fk we can decide whether G contains H as a minor in timeO(f(|H|)n3) for some function f : N → N. Let h be the size of the largest graphin Fk. Then deciding whether a particular graph G is in Ck can be done in timeO(|Fk|f(h)n3).

For a fixed minor closed graph class G, consider the following problem. Inputis graph G and integer k. The parameter is k and the objective is to determinewhether there exists a set vertex set S of size at most k such that G\S ∈ G. Theproof of Corollary 2.5.3 can easily be modified to show that for any fixed minorclosed graph class G this problem is FPT.

2.6 Win / Wins

The concept of Win/Wins is simple, yet it has proven useful both for showingproblems fixed parameter tractable and for designing efficient FPT algorithms.A Win / Win consists of perfoming a polynomial time test on the input instance,

23

and then proceeding depending on the test outcome. The idea is that for sometests all outcomes give useful information about the instance in question. Wegive an example with Max Leaf Spanning Tree (MLST). In this problemwe are given a graph G together with an integer k and asked whether there is aspanning tree of G with at least k leaves.

Theorem 2.6.1 ([58]) Max Leaf Spanning Tree is fixed parameter tractable.

Proof. On an input instance (G, k) we do a polynomial test. In particular wepick an arbitrary vertex v ∈ V (G) and run a breadth first search (BFS) startingat v. There are three outcomes, (a) : G is disconnected. (b) : G is connected,some BFS-layer has at least k vertices or (c) G is connected and all BFS layershave at most k−1 vertices. We show how all three outcomes give us informationthat help us decide the instance.

If G is disconnected then G has no spanning tree, so we can immediatelyanswer no. Assume now that G is connected. If some BFS-layer has at least kvertices then the BFS-tree rooted at v is a spanning tree with at least k leavesand we can answer yes. If all BFS layers have at most k−1 vertices, we constructa path decomposition of G my making a bag out of every two consequtive layers.This decomposition has width at most 2k − 3 so the pathwidth of G is at most2k − 3 in this case. Since the property that G has at least k leaves is express-ible in monadic second order logic, the theorem follows by Courcelle’s theorem.Expressing that G has at least k leaves in monadic second order logic is routine.

2.6.1 Bidimensionality

Win / Wins have been especially useful to develop subexponential time fixed pa-rameter algorithms for problems in planar graphs, and more generally in graphsexcluding a fixed graph H as a minor. For brevity, we only describe how these al-gorithms work in planar graphs. Many parameterized problems are optimizationproblems parameterzied by the objective function value. The objective functionis said to be bidimensional, if the optimal value of an n×m-grid is O(nm), andif the optimal value of any minor H of G is at most the optimal value of G.A parameterized problem is bidimensional if the input is a graph G and an in-teger k which is the parameter, and the objective is to determine whether theoptimal value of a bidimensional objective function on G is at most k. Mostbidimensional problems exhibit nice algorithmic properties on planar graphs. Inparticular, it was proved by Demaine et al. [43] that any bidimensional prob-lem which can be solved in time 2O(t+k)nO(1) on graphs of treewidth t admits a2O(

√k)nO(1) time algorithm in planar graphs. The results proved in [43] are for

more general classes of graphs and more general classes of problems than consid-ered here. We demonstrate how these ideas apply to Vertex Cover in planar

24

graphs. We will need two propositions about about treewidth and graph minorsthat we will use as black boxes. Notice that if a minor of G has treewidth at leastt then the treewidth of G is also at least t.

Proposition 2.6.2 (Excluded Grid Theorem [123]) Every planar graph Gof treewidth at least t contains a t

4× t

4-grid as a minor.

Proposition 2.6.3 ([125]) The treewidth of planar graphs can be 32-approximated

in polynomial time.

We explain how to exploit the above propositions to show that VertexCover in planar graphs can be solved in time 2O(

√k). As observed in Section 2.5,

if a graph has a vertex cover of size at most k, then so do all its minors. Observealso that the size of a minimum vertex cover of a k × k grid is at least k2/2.

Theorem 2.6.4 ([43, 46, 49]) There is an 2O(√

k)nO(1) time algorithm for Ver-tex Cover on planar graphs.

Proof. On an input instance (G, k) we perform a test in polynomial time - we runthe 3/2-approximation for treewidth on G, and let t be the width of the decomo-sition returned by the approximation algorithm. We have two possible outcomes,either ( 2t

3∗4)2/2 = t2

72> k or not. If not, then we can find an optimal vertex cover

in time O(2tnO(1)) ≤ 2O(√

(k)nO(1) by applying the 2tnO(1) time dynamic program-ming algorithm for Vertex Cover in graphs of bounded treewidth []. If t2

72> k

then G contains a (√

2k + 1×√

2k + 1) grid as a minor. The size of the minimumvertex cover of this grid is more than k and hence (G, k) is a no-instance.

Algorithms of the above type can be given for any bidimensional problemwhich can be solved efficiently in grahps of bounded treewidth, and several surverypapers have been written on Bidimensionality [43, 46, 49].

2.7 Color Coding

The Color Coding technique was developed by Alon et al [11] as a tool to detect ak-sized subgraph F of constant treewidth in an input graph G in time 2O(k)nO(1).The color coding technique is typically applied when we are searching for a smallstructure S in a larger structure X. We randomly color X with a set of colorsand show that if X contains S as a substructure then the probability that S hasbeen colored “nicely” is high. Here “nicely” means that the coloring is helpful forfinding S quickly. In this section we describe the color coding algorithm of Alonet al [11] for the Longest Path problem. In this problem we are given as inputa graph G and an integer k and asked whether G contains a path on k verticesas a subgraph.

25

We start by giving a randomized algorithm for the problem, and then showhow to derandomize it in order to get a deterministic algorithm. Our randomizedalgorithm will run in time O((2e)knO(1)) and if G does not contain a k-path thenthe algorithm returns that it does not. On the other hand, of G does contain ak-path the algorithm detects such a path with probability at least 3/4. Repeatingthe algorithm more times gives an arbitrarily good probability of returning thecorrect answer.

The algorithm proceeds as follows. We color V (G) with colors from 1, . . . , kuniformly at random, and run an algorithm to detect whether there is a pathP on k the vertices of P have been distinctly colored. We say that a path ismulticolored if it contains at most one vertex of each color. If G contains a pathP ′ on k vertices then the probability that P is multicolored is k!/kk = (1/ek)kO(1)

by Stirlings formula.

Lemma 2.7.1 ([11]) There is an O(2knO(1)) time algorithm to decide whetherthere is a multicolored path in a colored graph G.

Proof. Let V1, . . . , Vk be a partitioning of V (G) such that all vertices in Vi arecolored i. We apply dynamic programming, for a subset S of 1, . . . , k and avertex u /∈ ⋃i∈S Vi we define the function PATH(S, u) to return TRUE if thereis a multicolored path on |S| + 1 vertices in G[u⋃i∈S Vi] with one endpoint inu, and FALSE otherwise. Since any multicolored path can visit every color classat most once the recurrence

PATH(S, u) =∨

i∈S,v∈Vi,uv∈E(G)

PATH(S \ i, v)

holds. Clearly all values of PATH can be computed in O(2knO(1)) time by apply-ing the above recurrence. To determine whether there is a multicolored k-pathin G we loop over every i from 1 to k and for each vertex in Vi check whetherPATH(v, 1, . . . , i− 1, i+ 1, . . . , k) is set to TRUE. If it is for some choice fori and v, then there is a path, otherwise not. Given a positive answer the multi-colored path itself can be reconstructed using a standard backtracking technique.

Suppose G has a k-path. If we color V (G) with colors from 1, . . . , k uni-formly at random and then run the algorithm from Lemma 2.7.1 then the prob-ability that we detect this path is (1/(ekkO(1)). On the other hand, if G hasno k-path then we correctly return that it indeed does not. If we repeat thisprocedure O(ekkO(1)) times the probability that we detect a k-path, if it exists,is at least 3/4. This yields the O((2e)knO(1)) expected time algorithm for theLongest Path problem.

We now discuss how to derandomize the algorithm, that is, to replace therandomized coloring step by a deterministic step that does the same job. To thisend, the following result is helpful.

26

Proposition 2.7.2 ([114]) For every n, k there is a family of functions F ofsize O(ek · kO(log k) · log n) such that every function f ∈ F is a function from1, . . . , n to 1, . . . , k and for every subset S of 1, . . . , n there is a functionf ∈ F that is bijective when restricted to S. Furthermore, given n and k, F canbe computed in time O(ek · kO(log k) · logn).

The family F above can replace the random coloring step of our algorithm.In particular, instead of the random coloring step we construct the family F ,iterate over every f ∈ F and use f as a coloring of the graph. It follows fromProposition 2.7.2 that if G contains a path P on k vertices then there is somef ∈ F such that P is multicolored when f is used to color V (G). This yields thefollowing theorem.

Theorem 2.7.3 ([11, 114]) There is an O((2e)knO(1)) time algorithm for LongestPath.

For the case of Longest Path, algorithms with improved running times havelater been published. The currently fastest deterministic algorithm for the prob-lem runs in time O(4knO(1)) and is based on a combination of the Color Codingand Divide and Conquer techniques [95]. The fastest randomized algorithm isdue to Williams [136] and runs in expected time O(2knO(1)).

2.8 Integer Linear Programming

In the early 80’s, Lenstra [102] proved that Integer Linear Programmingis Fixed Parameter Tractable when parameterized by the number p of variables,giving an algorithm with running time doubly exponential in p. His result wassubsequently improved by Kannan [88] to a pO(p)nO(1) time algorithm. This algo-rithm uses Minkowski’s Convex Body theorem and other results from Geometryof Numbers. A bottleneck in this algorithm was that it required space exponen-tial in p. Using the method of simultaneous Diophantine approximation, Frankand Tardos [71] describe preprocessing techniques, using which it is shown thatLenstra’s and Kannan’s algorithms can be made to run in polynomial space.They also slightly improve the running time of the algorithm. For our purposes,we will use this algorithm.

Because of the vast expressive power of Integer Linear Programming it isnatural to expect that it should be possible to prove problems Fixed ParameterTractable by reducing the problem to Integer Linear Programming with fewvariables. Quite surprisingly this technique has not yet found many applicationsin Parameterized Complezity. To the authors best knowledge the method has onlybeen used to give a FPT algorithm for the Closest String problem [75] and,in an EPTAS for Min-Makespan-Scheduling problem [113] and to prove that

27

four graph layout problems are classFPT when parameterized by the size of theminimum vertex cover of the input graph [63]. We believe that it is quite probablethat Integer Linear Programming could turn out useful for other problems as well.In this section we state the theorems necessary to apply the method and give anexample of an FPT algorithm based on Integer Linear Programming.

Integer Linear Programming (p-ILP): Given matrices A ∈ Zm×p

and b ∈ Zm×1, the question is whether there exists a vector x ∈ Z

p×1

satisfying the m inequalities, that is, A · x ≤ b. The number of vari-ables p is the parameter.

Theorem 2.8.1 ([88],[102],[71]) p-ILP can be solved using O(p2.5p+o(p) · L)arithmetic operations and space polynomial in L. Here L is the number of bits inthe input and p the number of variables in the integer linear program.

In order to be able to use an algorithm for p-ILP as a subroutine in ouralgorithms, we need the optimization version of p-ILP rather than the feasibilityversion. We proceed to define the minimization version of p-ILP.

p-Variable Integer Linear Programming Optimization (p-Opt-ILP): Let matrices A ∈ Z

m×p, b ∈ Zm×1 and c ∈ Z

1×p be given.We want to find a vector x ∈ Z

p×1 that minimizes the objectivefunction c · x and satisfies the m inequalities, that is, A · x ≥ b. Thenumber of variables p is the parameter.

Theorem 2.8.2 p-Opt-ILP can be solved using O(p2.5p+o(p)·L·log (MN)) arith-metic operations and space polynomial in L. Here, L is the number of bits in theinput, N is the maximum of the absolute values any variable can take, and Mis an upper bound on the absolute value of the minimum taken by the objectivefunction.

Proof. We can first do a binary search to find the minimum value of the objectivefunction. For an example suppose we guess that c · x ≤ M

2. Now we make a new

matrix A′ of dimension (m + 1) × p whose first row consists of c and rest of therows is −A(by multiplying each entry of A by −1). We will denote the matrix

A′ as [ c−A

]. Similarly we make b′ = [M/2−b

]. Now we apply Theorem 2.8.1 to checkwhether there exists a vector x′ ∈ Z

p×1 such that A′ · x′ ≤ b′. Having found theminimum value of the objective function in this way, we find the lexicographicallysmallest solution satisfying these inequalities. We determine the value of onevariable at a time by doing a binary search similar to the one we used to find theminimum of the objective function. For all this we need to run the algorithm forILP easibility at most O(p · logN + logM) times.

Now we give an example for how Theorem 2.8.2 can be used applied to showthat a problem is FPT.

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An Algorithm for Graph Imbalance: In order to define the Imbalanceproblem we need to introduce some notation. A permutation π : V → 1, 2, . . . , norders the vertex set into v1 <π v2 <π . . . <π vn. For every i, the set Vi isv1, . . . , vi and ∂(Vi) = uv | uv ∈ E, u ∈ Vi, v ∈ V \ Vi. We define Lπ(v) to beu | u ∈ N(v), u <π v and Rπ(v) is u | u ∈ N(v), v <π u.

In the Imbalance problem, we are given a graph G = (V,E) with a vertexcover C = c1, . . . , ck ⊆ V of size k as input and asked to find a permutationπ : V → 1, 2, . . . , n that minimizes

∑i≤n |Lπ(vi)−Rπ(vi)|. The parameter is k,

that is, the size of the vertex cover C. One should notice that this means that analgorithm whose running time depends exponentially on the objective function isnot and FPT algorithm for this parameterized problem. We show how to applyTheorem 2.8.2 in order to give an FPT algorithm for Imbalance parameterizedby the size of the minimum vertex cover of G.

We are looking for a permutation π : V → 1, 2, . . . , n for which fim(π) isminimized. In order to do this, we loop over all possible permutations of thevertex cover C and for each such permutation πc, find the best permutationπ of V that agrees with πc. We say that π and πc agree if for all ci, cj ∈ Cwe have that ci <π cj if and only of ci <πc

cj . In other words, the relativeordering π imposes on C is precisely πc. Thus, at a cost of a factor of k! in therunning time we can assume that there exists an optimal permutation π suchthat c1 <π c2 <π . . . <π ck.

Definition 2.8.3 Let πc be an ordering of C such that c1 <πcc2 <πc

. . . <πcck.

We define Ci to be c1, c2, . . . , ci for every i such that 1 ≤ i ≤ k.

Let I be the independent set V \ C. We associate a type with each vertex inI. A type is simply a subset of C.

Definition 2.8.4 Let I be the independent set V \C. The type of a vertex v inI is N(v). For a type S ⊆ C the set I(S) is the set of all vertices in I of type S.

Notice that two vertices of the same type are indistinguishable up to automor-phisms of G, and that there are 2k different types.

Observe that every vertex of I is either mapped between two vertices of C, tothe left of c1 or to the right of ck by a permutation π. For a permutation π wesay that a vertex v is at location 0 if v <π c1 and at location i if i is the largestinteger such that ci <π v. The set of vertices that are at location i is denoted byLi. We define the inner order of π at location i to be the permutation defined byπ restricted to Li.

The task of finding an optimal permutation can be divided into two parts.The first part is to partition the set I into L0, . . . , Lk, while the second partconsists of finding an optimal inner order at all locations. One should notice thatpartitioning I into L0, . . . , Lk amounts to deciding how many vertices of eachtype are at location i for each i. Observe that permuting the inner order of π at

29

location i does not change the imbalance of any single vertex, where the imbalanceof a vertex v is |Lπ(v) − Rπ(v)|. Hence, the inner orders are in fact irrelevantand finding the optimal ordering of the vertices thus reduces to finding the rightpartition of I into L0, . . . , Lk. We formalize this as an instance of p-Opt-ILP.

For a type S and location i we let xiS be a variable that encodes the number

of vertices of type S that are at location i. Also, for every vertex ci in C we havea variable yi that represents the imbalance of ci. In order to represent a feasiblepermutation, all the variables must be non-negative. Also the variables xi

S mustsatisfy that for every type S,

∑ki=0 x

iS = |I(S)|. For every vertex ci of the vertex

cover let ei = |N(ci) ∩Ci−1| − |N(ci) ∩ (C \ Ci)| be a constant. Finally for everyci ∈ C we add the constraint

yi =∣∣ei +

∑

S⊆C|ci∈S

( i−1∑

j=0

xjS −

k∑

j=i

xjS

)∣∣.

One should notice that the last set of constraints is not a set of linear con-straints. However, we can guess the sign of

y′i = ei +∑

S⊆C|ci∈S

( i−1∑

j=0

xjS −

k∑

j=i

xjS

)

for every i in an optimal solution. This increases the running time by a factor of2k. For every i we let ti take the value 1 if we have guessed that y′i ≥ 0 and welet ti take the value −1 if we have guessed that y′i < 0. We can now replace thenon-linear constraints with the linear constraints yi = tiy

′i for every i. Finally,

for every type S and location i, let ziS be the constant

∣∣|S ∩Ci| − |S ∩ (C \Ci)|∣∣.

We are now ready to formulate the integer linear program.

min

k∑

i=1

ti · yi +∑

S⊆C

ziS · xi

S

such that∑

i

xiS = |I(S)| for all i ∈ 0, . . . , k, S ⊆ C

yi = tiei +∑

S⊆C|ci∈S

(∑i−1j=0 tix

jS −∑k

j=i tixjS

)for all i ∈ 1, . . . , k

xiS , yi ≥ 0 for all i ∈ 0, . . . , k, S ⊆ C

Since the value of fim(π) is bounded by n2 and the value of any variable in theinteger linear program is bounded by n, Theorem 2.8.2 implies that this integerlinear program can be solved in FPT time, thus implying the following theorem.

Theorem 2.8.5 The Imbalance problem parameterized by the vertex covernumber of the input graph is fixed parameter tractable.

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The proof of Theorem 2.8.5 serves as an example for how to apply Theorem2.8.2 to give FPT algorithms. The result of Lenstra was extended by Khachiyanand Porkolab [92] to semidefinite integer programming. In their work, they showthat if Y is a convex set in Rk defined by polynomial inequalities and equations ofdegree at most d ≥ 2, with integer coefficients of binary length at most l, then forfixed k, the problem of computing an optimal integral solution y∗ to the problemmin yk | y(y1, . . . , yl) ∈ Y ∪Zk admits an FPT algorithm. Their algorithm wasfurther improved by Heinz [84] in the specific case of minimizing a polynomial Fon the set of integer points described by an inequality system Fi ≤ 0, 1 ≤ i ≤ swhere the Fi’s are quasiconvex polynomials in p variables with integer coefficients.It would be interesting to see if these even more general results can be useful forshowing problems fixed parameter tractable.

Chapter 3

Running Time Lower Bounds

In this chapter we discuss techniques for showing lower bounds for the runningtime of algorithms for parameterized problems, up to certain complexity theoreticassumptions. We take an “engineering” approach to the topic, that is, we do notdwell on the underlying complexity theory. Instead we discuss the types of boundsone can obtain under different assumptions and give examples of how to showsuch bounds.

3.1 NP-hardness vs. ParaNP-hardness

NP-hardness of a parameterized problem rules out algorithms with running timenO(1), assuming P 6= NP . On the other hand, one might have an algorithm withrunning time nf(k). For instance, the Independent Set problem admits a simpleO(nk) time algorithm: try all possible vertex subsets of size k and check whetherthey form an independent set. However, not all problems have algorithms withrunning time of this form. Some parameterized problems, as we demonstratebelow, remain NP-hard even when k is a fixed integer. Parameterized problemsthat are NP-hard for every fixed value of k above some threshold K are calledParaNP-hard. For these problems one can not hope for algorithms with runningtime on the form O(nO(f(k))), let alone FPT algorithms.

Theorem 3.1.1 ([126]) Graph Coloring is ParaNP-hard

Proof. We show that for every fixed k ≥ 3, the problem of deciding whetheran input graph G is k-colorable is NP-complete. A reduction showing thatGraph Coloring is NP-complete for k = 3 is given in for example the bookof Sipser [126]. For a given graph G we can make a graph G′ by adding a newvertex adjacent to all vertices of G. It follows that G is k-colorable if and onlyif G′ is k + 1-colorable and hence Graph Coloring is NP-complete for everyfixed k > 3.

31

32

Quite a few problems share this property, and for many of them showingParaNP-hardness turned out to be easier than showing W [1]-hardness (explainedbelow). Examples include Treelength [103], Metric Embedding Into theLine [60] and Minimum Fill-in parameterized by the number of vertices thathas to be deleted from the input graph to make it chordal [106].

3.2 Hardness for W[1]

How can we show that a problem does not have an algorithm with running timef(k)·nO(1)? One approach would be to show NP-hardness. But with this method,we can not distinguish between problems that are solvable in time nf(k) from prob-lems solvable in time f(k) ·nO(1). To be able to do this, Downey and Fellows [52]introduced the W-hierarchy. The hierarchy consists of a complexity class W[t]for every integer t ≥ 1 such that W[t] ⊂ W[t + 1] for all t. Downey and Fel-lows [52] proved that FPT ⊆ W[1] ⊆ W[2] . . . ⊆ W[t] and conjectured that strictcontainment holds.

In particular, the assumption FPT ⊂ W[1] is the fundamental complexitytheoretic assumption in Parameterized Complexity. The reason for this is that theassumption is a natural parameterized analogue of the conjecture that P 6= NP .The assumption P 6= NP can be reformulated as “The Non-DeterministicTurning Machine Acceptance problem can not be solvable in polynomialtime”. In this problem we are given a non-deterministic turing machine M , astring s and an integer k coded in unary. The question is whether M can makeits non-deterministic choices in such a way that it accepts s in at most k steps.The intuition behind the P 6= NP conjecture is that this problem is so generaland random that it is not likely to be in P. Similarly, one would not expect theproblem parameterized by k to be fixed parameter tractable.

From our “engineering” viewpoint, how can we use the assumption thatFPT 6= W[1] to rule out f(k) · nO(1) time algorithms for a particular problem?Downey and Fellows showed that the Independent Set problem is not in FPTunless FPT = W[1]. If we can show for a particular parameterized problem Π,that if Π is fixed parameter tractable then so is Independent Set, then thisimplies that Π /∈ FPT unless FPT = W[1]. To this end, we define the notion ofFPT-reductions.

Definition 3.2.1 ([52]) Let P and Q be parameterized problems. We say thatP is FPT-reducible to Q, written P ≤FPT Q, if there exists an algorithm thatgiven as input an instance (x, k), runs in time |x|O(1)f(k) for some functionf : N :→ N, and outputs an instance (x′, k′) such that (a) (x, k) ∈ P if and only(x′, k′) = f(x, k) ∈ Q and (b) k′ ≤ g(k) for some function g : N :→ N.

The class W[1] is the set of all parameterized problem that are FPT-reducibleto the parameterized Non-Deterministic Turning Machine Acceptance

33

problem. A parameterized problem is said to be W[1]-hard if all problems in W[1]FPT-reduce to it. The following theorem follows directly from Definition 3.2.1.

Theorem 3.2.2 ([52]) Let P and Q be parameterized problems. If P ≤FPT Qand Q is in FPT then P is in FPT. Furthermore, if P ≤FPT Q and P is W[1]-hard then Q is W[1]-hard.

Theorem 3.2.2 implies that a W[1]-hard problem can not be in FPT unlessFPT = W[1]. As mentioned above, it was proved in [52] that IndependentSet is W[1]-hard. We now give two examples of FPT-reductions. First, we showthat the Multicolor Clique problem is W[1]-hard.

Multicolor Clique: Given an integer k and a connected undi-rected graph G = (V [1] ∪ V [2] · · · ∪C[k], E) such that for every i thevertices of V [i] induce an independent set, is there a k-clique C in G?

The approach for using the Multicolor Clique problem in reductions isdescribed in [62], and has been proven to be very useful in showing hardnessresults in Parameterized Complexity.

Theorem 3.2.3 Multicolor Clique is W[1]-hard.

Proof. We reduce from the Independent Set problem. Given an instance(G, k) to Independent Set we construct a new graph G′ = (V ′, E ′) as follows.For each vertex v ∈ V (G) we make k copies of v in V ′ with the i’th copy beingcolored with the i’th color. For every pair u,v ∈ V (G) such that uv /∈ E(G) weadd edges between all copies of u and all copies of v with different colors. It iseasy to see that G has an independent set of size k if and only if G′ contains aclique of size k. This concludes the proof.

One should notice that the reduction produces instances to MulticolorClique with a quite specific structure. In particular, all color classes have thesame size and the number of edges between every pair of color classes is the same.It is often helpful to exploit this fact when reducing from Multicolor Cliqueto a specific problem. We now give an example of a slightly more involved FPT-reduction.

Theorem 3.2.4 Dominating Set is W[1]-hard.

Proof. We reduce from the Multicolor Clique problem. Given an instance(G, k) to Multicolor Clique we construct a new graph G′. For every i ≤ klet Vi be the set of vertices in G colored i and for every pair of distinct integersi, j ≤ k let Ei,j be the set of edges in G[Vi ∪ Vj ]. We start making G′ by takinga copy of Vi for every i ≤ k and making this copy into a clique. Now, for every

34

i ≤ k we add a set Si of k + 1 vertices and make them adjacent to all verticesof Vi. Finally, for every pair of distinct integers i, j ≤ k we consider the edgesin Ei,j.For every pair of vertices u ∈ Vi and v ∈ Vj such that uv /∈ Ei,j we adda vertex xuv and make it adjacent to all vertices in Vi \ u and all vertices inVj \ v. This concludes the construction. We argue that G contains a k-cliqueif and only if G′ has a dominating set of size at most k.

If G contains a k-clique C then C is a dominating set of G′. In the otherdirection, suppose G′ has a dominating set S of size at most k. If for some i,S ∩ Vi = ∅ then Si ⊆ S, contradicting that S has size at most k. Hence for everyi ≤ k, S∩Vi 6= ∅ and thus S contains exactly one vertex vi from Vi for each i, andS contains no other vertices. Finally, we argue that S is a clique in G. Supposethat vivj /∈ Ei,j . Then there is a vertex x in V (G′) with neighbourhood Vi \ uand Vj \ v. This x is not in S and has no neighbours in S contradicting that Sis a dominating set of G′.

In fact, it was an FPT-reduction from Independent Set to DominatingSet that was the startpoint of the complexity part of parameterized algorithmsand complexity. In 1989, Fellows realised that one could give a FPT-reductionfrom Independent Set to Dominating Set, but that it did not seem plausibleto give a reduction in the other direction. Later, Downey and Fellows provedthat the Dominating Set problem in fact is complete for the class W[2] whileIndependent Set is complete for W[1] [52]. It is therefore unlikely that anFPT-reduction from Dominating Set to Independent Set can exist. Thetheorem 3.2.4 is due to Downey and Fellows. The proof presented in this thesis issomewhat simpler than the original proof and due to the author. For a thoroughintroduction to the W-hierarchy we refer the reader to the books of Downey andFellows [52] and Flum and Grohe [66].

3.3 Hardness for W[2]

Even though it was shown by Downey and Fellows that Dominating Set iscomplete for W[2] while Independent Set is complete for W[1] [52], there areproblems for which it is easier to give FPT-reductions from Dominating Setthan from Independent Set. A typical example is the Steiner Tree problem,parameterized by the number of non-terminals in the solution. In the SteinerTree problem we are given a connected graph G together with a subset X ofV (G) and an integer k. The vertices in the X are called terminals. The objectiveis to find a subtree T of G containing all terminals and at most k non-terminals.Such a tree is called a steiner tree of G. Here we give a proof that the SteinerTree problem parameterized by the number of non-terminals in the steiner treeis W[2] hard. The proof of the following theorem first appeared in [17].

Theorem 3.3.1 ([17]) The Steiner Tree problem parameterized by |V (T ) \

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X| is W[2]-hard.

Proof. We reduce from the Dominating Set problem. For an instance (G, k)we build a graph G′ as follows. We make two copies of V (G), call them X ′ andN . The copy x ∈ X ′ of each vertex v ∈ V (G) is made adjacent to the copies in Nof the vertices in N(v). Finally the vertex set X is obtained from X ′ by addinga single vertex uX and making it adjacent to all vertices in N . This concludesthe construction of G′. We prove that G has a dominating set of size at most kif and only if G′ has a steiner tree with at most |X| + k vertices.

In one direction, suppose G has a dominating set S on k vertices. Let S ′ bethe copy of S in N . Then the graph G′[X ∪S ′] is connected. Let T be a spanningtree of G′[X ∪S ′], then T is a steiner tree of G′ with at most |X|+ k vertices. Inthe other direction, suppose G′ has a steiner tree T on at most |X| + k verticesand let S ′ = V (T ) ∩ N . Then |S ′| ≤ k and since X is an independent set in G′

every vertex in X has a neighbour in S ′. Thus, if we let S be the copy of S ′ inV (G) then S is a dominating set of G of size at most k.

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Chapter 4

Kernelization

Polynomial time preprocessing to reduce instance size is one of the most widelyused approaches to tackle computationally hard problems. A natural questionin this regard is how to measure the quality of preprocessing rules proposed fora specific problem. Parameterized complexity provides a natural mathematicalframework to give performance guarantees of preprocessing rules. A parame-terized problem is said to admit an f(k) kernel if there is a polynomial timealgorithm, called a kernelization algorithm, that reduces the input instance downto an instance with size bounded by f(k) in k, while preserving the answer. Thisreduced instance is called an f(k) kernel for the problem.

Definition 4.0.2 A kernelization algorithm, or in short, a kernel for a parame-terized problem Q ⊆ Σ∗ × N is an algorithm that given (x, k) ∈ Σ∗ × N outputsin time polynomial in |x| + k a pair (x′, k′) ∈ Σ∗ × N such that (a) (x, k) ∈ Qif and only if (x′, k′) ∈ Q and (b) |x′|, k ≤ g(k), where g is an arbitrary com-putable function. The function g is referred to as the size of the kernel. If g is apolynomial function then we say that Q admits a polynomial kernel.

It is easy to see that if a decidable problem admits an f(k) kernel for somefunction f , then the problem is FPT. Interestingly, the converse also holds, thatis, if a problem is FPT then it admits an f(k) kernel for some function f [116].The proof of this fact is quite simple, and we present it here.

Fact 4.0.3 (Folklore, [116]) If a parameterized problem Π is FPT then Π ad-mits an f(k) kernel for some function f .

Proof. Suppose there is a decision algorithm for Π running in f(k)nc time forsome function f and constant c. Given an instance (I, k) with |I| = n, if n ≥ f(k)then we run the decision algorithm on the instance in time f(k)nc ≤ nc+1. If thedecision algorithm outputs yes, the kernelization algorithm outputs a constantsize yes instance, and if the decision algorithm outputs no, the kernelization

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algorithm outputs a constant size no instance. On the other hand, if n < f(k)the kernelization algorithm just outputs (I, k). This yields an f(k) kernel for theproblem.

Fact 4.0.3 implies that a problem has a kernel if and only if it is fixed parametertractable. However, we are interested in kernels that are as small as possible, anda kernel obtained using Fact 4.0.3 has size that equals the dependence on k in therunning time of the best known FPT algorithm for the problem. The questionis - can we do better? The answer is that quite often we can. In fact, for manyproblems we can obtain kc kernels, called polynomial kernels. In this chapter wesuvey some of the known techniques for showing that problems admit polynomialkernels.

4.1 Reduction Rules

We have already seen the main tool for showing that a problem admits an f(k)-kernel in a different context. Namely, we briefly touched upon reduction rulesin Section 2.2. Recall that a reduction rule is a polynomial time algorithm thattransforms an instance (I, k) of a parameterized problem to an ‘equivalent andsimpler instance (I ′, k′), where equivalent means that (I, k) is a yes instance ifand only if (I ′, k′) is. In Section 2.2 we used reduction rules to identify “good”structures to branch on. Here we will show that for some problems we can givereduction rules such that any instance to which no reduction rules can be appliedmust have size bounded by a function of k. This will immediately give us thedesired kernel.

We show how a simple set of reduction rules gives a k2 kernel for the PointLine Cover problem. An instance of this problem is a pair (S, k) where S is aset of n points in the plane with no two points in the exact same spot and k isan integer. The objective is to find a set C of k lines in the plane such that everypoint lies on some line in C. For a line l in the plane let P (l) be the set of pointsthat are covered by l. To tackle the problem, it is enough to make the followingobservations:

1. It is sufficient to only consider lines containing at least 2 points.

2. Any two lines intersect in at most one point.

Hence any line that contains at least k + 1 points must be in the solution setS, since otherwise S must contain at least k + 1 lines. This yields the followingreduction rule.

RedRule 4.1.0.1 If some line l covers at least k+1 points, return (S \P (l), k−1).

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Rule 4.1.0.1 is correct since any line except for l covers at most one point inP (l). Hence if the solution set C does not contain l then it must contain at leastP (l) > k lines.

Theorem 4.1.1 The Point Line Cover problem admits a k2 kernel.

Proof. Consider an instance (S, k) to which Rule 4.1.0.1 can not be applied. Weargue that if |S| > k2, then (S, k) is a no-instance. Suppose for contradictionthat there is a set C of at most k lines such that the lines in C together coverall points in S. Since Rule 4.1.0.1 can not be applied every line in C covers atmost k points in S. Thus the lines in C cover at most k2 points of S in total,contradicting that S is covered.

Theorem 4.1.1 imediately yields a O(k4knO(1)) time algorithm for the Point

Line Cover problem: after obtaining the k2 kernel, iterate over all((k2

2 )k

)ways to

chose k lines and check whether they cover all the points. In fact, a O(k2) kerneland a O(kO(k)nO(1)) time algorithm for this problem is a corollary of a result byLangerman and Morin [100]. However, no o(k2) sized kernel or O(2o(k2)nO(1))time algorithm is known for the problem to this date and remains an interestingopen problem.

4.2 Crown Reductions

A crown decomposition of a graph G is a decomposition of V (G) into C ⊎H ⊎Rsuch that N(C) ⊆ H , G[C] is an independent set and there is a matching from Hto C. The set C is called the crown, H is called the head and R is called the restor body of the decomposition. For many problems, if a crown decomposition of theinput graph, or a graph modelling the problem, is found then the crown can beremoved or reduced. We give an example by giving a kernel for the MaximumSatisfiability problem. In the Maximum Satisfiability problem we aregiven a formula φ in conjunctive normal form and an integer k. The formula hasn variables and m clauses and the question is whether there is an assignmentto the variables that satisfies at least k of the clauses. We give a kernel basedon crown reductions with less than k variables and less than 2k clauses. LetGφ be the variable-clause incidence graph of φ. That is, Gφ is a bipartite graphwith one bipartition X corresponding to the variables of φ and a bipartition Ycorresponding to the clauses. For a vertex x ∈ X we will refer to x as both thevertex in Gφ and the corresponding variable in φ. Similarly, for a vertex c ∈ Y wewill refer to c as both the vertex in Gφ and the corresponding clause in φ. In Gφ

there is an edge between a variable x ∈ X and a clause c ∈ Y if x or its negationoccurs in c. We assume that every clause of φ contains at least one variable.

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Theorem 4.2.1 The Maximum Satisfiability problem admits a kernel withless than k variables and less than 2k clauses.

Proof. If we assign values to the variables uniformly at random, linearity ofexpectation yields that the expected number of satisfied clauses it at least m/2.Since there has to be at least one assigment satisfying at least the expectednumber of clauses this means that if m ≥ 2k then (φ, k) is a yes-instance. In therest of this section we show how to give a kernel where n < k. Whenever possiblewe apply a cleaning rule; if some variable does not occur in any clauses, removethe variable.

Let Gφ be the variable-clause incidence graph of φ, X and Y be the bipar-titions of Gφ corresponding to variables and clauses respectively. If there is amatching from X to Y in Gφ there is an assignment to the variables satisfyingat least n clauses. This is true because we can set each variable in X in such away that the clause matched to it becomes satisfied. In this manner, at least |X|clauses are satisfied. We now show that if φ has at least k variables then we can,in polynomial time, either reduce φ to an equivalent smaller instance or find anassignment to the variables satisfying at least k clauses.

Suppose φ has at least k variables. Using Hall’s Theorem and an algorithmfor Maximum Matching we can in polynomial time find either a matching fromX to Y or an inclusion minimal set C ⊆ X such that |N(C)| < |C|. If we found amatching we are done, as we can satisfy at least |X| ≥ k clauses. So suppose wefound a set C as described. Let H be N(C) and R = V (Gφ) \ (C ∪H). Clearly,N(C) ⊆ H , N(R) ⊆ H and G[C] is an independent set. Furthermore, for a vertexx ∈ C we have that there is a matching from C \ x to H since |N(C ′)| ≥ |C ′|for any C ′ ⊆ (C \ x). Since |C > H| the matching from C \ x to H is in fact amatching from H to C. Hence C ⊎H ⊎ R is a crown decomposition of Gφ.

We prove that all clauses in H are satisfied in every truth assignment tothe variables satisfying the maximum number of clauses. Indeed, consider anyassignment ψ of values to the variables that does not satisfy all clauses in H . Forevery variable y in C\x change the value of y such that the clause inH matchedto y is satisfied. Let ψ′ be the new assignment obtained from ψ in this manner.Since N(C) ⊆ H and ψ′ satisfies all clauses in H , ψ′ satisfies more clauses thanψ does, and hence ψ can not be an assigment satisfying the maximum number ofclauses.

The argument above shows that (φ, k) is a yes instance to Maximum Sat-isfiability if and only if (φ \ (C ∪ H), k − |H|) is. This gives rise to a simplereduction rule: remove (C ∪ H) from φ and decrease k by |H|. This completesthe proof of the theorem.

To the author’s best knowledge, the best kernel known before this result is akernel with 2k variables and O(k2) clauses [116].

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4.3 LP based kernelization

One of the classical results in algorithms is a polynomial time algorithm to solveLinear Programming [93]. This result has proved extremely useful in thedesign of approximation algorithms. In particular, any problem in NP can beformulated as an integer linear program (ILP), an LP with the additional restric-tion that the variables only are allowed to take integer values. Since IntegerLinear Programming is NP-complete we can not hope to solve the integer lin-ear program in polynomial time. Instead we settle for an approximate solution bysolving the corresponding linear programming relaxation of the ILP, which is justthe ILP without the integrality constraint on the variables. A (not necessariallyoptimal) solution to the ILP is obtained by rounding the variables in an optimalsolution to the LP relaxation in an appropriate way. Linear Programming isalso helpful to give problem kernels. We show how to obtain a 2k kernel for theVertex Cover problem by applying Linear Programming. This kernel is dueto [29, 115].

Given a graph G and integer k we construct an ILP with n variables, onevariable xv for each vertex v ∈ V (G). Setting the variable xv to 1 means that vgoes into the vertex cover, while setting xv to 0 means that v is not in the vertexcover. This yields the following ILP:

min∑

v∈V (G)

xv subject to:

xu + xv ≥ 1 for every uv ∈ E(G)

xv ∈ 0, 1 for every v ∈ V (G)

Clearly the optimal value of the ILP is at most k if and only if G has a vertexcover of size at most k. We relax the ILP by replacing the constraint xv ∈ 0, 1for every v ∈ V (G) with the constraint 0 ≤ xv ≤ 1 for every v ∈ V (G). Nowwe solve this LP in polynomial time. If the optimal value of the LP is greaterthen k then it must be greater than k for the ILP as well and we return thatG has no vertex cover of size at most k. Otherwise, let x∗v be the value of xv inthe produced optimal solution to the LP. We partition the vertex set of G intothree sets, C = v : x∗v < 1/2, H = v : x∗v > 1/2 and R = v : x∗v = 1/2.The name choice for the three sets is not arbitrary, as V (G) = C ⊎H ⊎ R is infact a crown decomposition of G. No edge can go between two vertices of C orbetween a vertex of C and a vertex of R since this would violate the constraintthat xu + xv ≥ 1 for every uv ∈ E(G). Furthermore, if for any subset H ′ ⊆ Hwe have |H ′| > |N(H)∩C| then decreasing the value of xv by ǫ for every v ∈ H ′

and increasing the value of x∗v by ǫ for every v ∈ N(H)∩C would yield a feasiblesolution to the LP with a lower value of min

∑v∈V (G) xv, contradictiong that x∗

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was an optimal solution to the LP. By Halls Theorem H is matched into C in G.For a vertex v ∈ H , let m(v) be the vertex in C matched to v.

Theorem 4.3.1 The Vertex Cover problem admits a 2k kernel.

Proof. We first prove that there is an optimal vertex cover S of G containing allvertices of H and no vertices of C. Let S ′ be an optimal vertex cover of G. Forevery vertex v ∈ H \ S ′, m(v) must be in S ′. Hence S = (S ′ ∪H) \C is a vertexcover of G with |S| ≤ |S ′|. This gives rise to a reduction rule since (G, k) is a yesinstance to Vertex Cover if and only if (G \ (C ∪ S), k − |H|) is. If H and Care non-empty, remove H and C from the graph and reduce k by |H|. Now, if Cis empty then x∗v ≥ 1/2 for all v ∈ V (G). But

∑v∈V (G) xv ≤ k so |V (G)| ≤ 2k.

Chapter 5

Kernelization Lower Bounds

In Chapter 4 we saw that a parameterized problem admits a kernel if and only ifit is FPT, and that for many fixed parameter tractable problems we are able toobtain kernels of size polynomial in the parameter k. In Chapter 3 we discussedmethods for showing that a problem is not FPT under different complexity-theoretic assumptions. Of course, due to Fact 4.0.3 if we can show that a prob-lem is not FPT under a certain complexity-theoretic assumption then under thesame assumption the problem does not admit an f(k) kernel for any function f .However, this method will not let us separate “good”, that is polynomial, kernelsfrom “bad” f(k) kernels. It is only very recently that a methodology to rule outpolynomial kernels has been developed [20, 70]. In this chapter we survey thetecniques that have been developed to show kernelization lower bounds.

5.1 Composition Algorithms

Consider the Longest Path problem. In Section 2.7 we saw an FPT algo-rithm for this problem. Is it feasible that it also admits a polynomial kernel?We argue that intuitively this should not be possible. Consider a large set(G1, k), (G2, k), . . . , (Gt, k) of instances to the Longest Path problem. If wemake a new graph G by just taking the disjoint union of the graphs G1 . . . Gt

we see that G contains a path of length k if and only if Gi contains a path oflength k for some i ≤ t. Suppose the Longest Path problem had a polynomialkernel, and we ran the kernelization algorithm on G. Then this algorithm wouldin polynomial time return a new instance (G′, k′) such that |V (G)| = kO(1), anumber potentially much smaller than t. This means that in some sense, thekernelization algorithm considers the instances (G1, k), (G2, k) . . . (Gt, k) and inpolynomial time figures out which of the instances are the most likely to containa path of length k. However, at least intuitively, this seems almost as difficultas solving the instances themselves and since the Longest Path problem isNP-complete, this seems unlikely. We now formalize this intuition.

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Definition 5.1.1 [Distillation [20]]

• An OR-distillation algorithm for a language L ⊆ Σ∗ is an algorithm thatreceives as input a sequence x1, . . . , xt, with xi ∈ Σ∗ for each 1 ≤ i ≤ t,uses time polynomial in

∑ti=1 |xi|, and outputs y ∈ Σ∗ with (a) y ∈ L ⇐⇒

xi ∈ L for some 1 ≤ i ≤ t and (b) |y| is polynomial in maxi≤t |xi|. Alanguage L is OR-distillable if there is a OR-distillation algorithm for it.

• An AND-distillation algorithm for a language L ⊆ Σ∗ is an algorithm thatreceives as input a sequence x1, . . . , xt, with xi ∈ Σ∗ for each 1 ≤ i ≤ t,uses time polynomial in

∑ti=1 |xi|, and outputs y ∈ Σ∗ with (a) y ∈ L ⇐⇒

xi ∈ L for all 1 ≤ i ≤ t and (b) |y| is polynomial in maxi≤t |xi|. A languageL is AND-distillable if there is an AND-distillation algorithm for it.

Observe that the notion of distillation is defined for unparameterized problems.Bodlaender et al. [20] conjectured that no NP-complete language can have anOR-distillation or an AND-distillation algorithm.

Conjecture 5.1.2 (OR-Distillation Conjecture [20]) No NP-complete lan-guage L is OR-distillable.

Conjecture 5.1.3 (AND-Distillation Conjecture [20]) No NP-complete lan-guage L is AND-distillable.

One should notice that if any NP-complete language is distillable, then soare all of them. Fortnow and Santhanam [70] were able to connect the OR-Distillation Conjecture to a well-known conjecture in classical complexity. Inparticular they proved that if the OR-Distillation Conjecture fails, the polynomialtime hierarchy [131] collapses to the third level, a collapse that is deemed unlikely.No such connection is currently known for the AND-Distillation Conjecture, andfor reasons soon to become apparent, a proof of such a connection would havesignificant impact in Parameterized Complexity. By PH=Σ3

p we will denote thecomplexity-theoretic event that the polynomial time hierarchy collapses to thethird level.

Theorem 5.1.4 ([70]) If the OR-Distillation Conjecture fails, then PH=Σ3p.

We are now ready to define the parameterized analogue of distillation algo-rithms and connect this notion to the Conjectures 5.1.2 and 5.1.3

Definition 5.1.5 [Composition [20]]

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• A composition algorithm (also called OR-composition algorithm) for a pa-rameterized problem Π ⊆ Σ∗ × N is an algorithm that receives as input asequence ((x1, k), . . . , (xt, k)), with (xi, k) ∈ Σ∗ × N

+ for each 1 ≤ i ≤ t,uses time polynomial in

∑ti=1 |xi|+k, and outputs (y, k′) ∈ Σ∗×N

+ with (a)(y, k′) ∈ Π ⇐⇒ (xi, k) ∈ Π for some 1 ≤ i ≤ t and (b) k′ is polynomialin k. A parameterized problem is compositional (or OR-compositional) ifthere is a composition algorithm for it.

• An AND-composition algorithm for a parameterized problem Π ⊆ Σ∗ × N

is an algorithm that receives as input a sequence ((x1, k), . . . , (xt, k)), with(xi, k) ∈ Σ∗×N

+ for each 1 ≤ i ≤ t, uses time polynomial in∑t

i=1 |xi|+ k,and outputs (y, k′) ∈ Σ∗ × N

+ with (a) (y, k′) ∈ Π ⇐⇒ (xi, k) ∈ Π forall 1 ≤ i ≤ t and (b) k′ is polynomial in k. A parameterized problem isAND-compositional if there is an AND-composition algorithm for it.

Composition and distillation algorithms are very similar. The main differencebetween the two notions is that the restriction on output size for distillationalgorithms is replaced by a restriction on the parameter size for the instancethe composition algorithm outputs. We define the notion of the unparameterizedversion of a parameterized problem L. The mapping of parameterized problemsto unparameterized problems is done by mapping (x, k) to the string x#1k, where# /∈ Σ denotes the blank letter and 1 is an arbitrary letter in Σ. In this way,the unparameterized version of a parameterized problem Π is the language Π =x#1k | (x, k) ∈ Π. The following theorem yields the desired connection betweenthe two notions.

Theorem 5.1.6 ([20, 70]) Let Π be a compositional parameterized problem whose

unparameterized version Π is NP-complete. Then, if Π has a polynomial kernelthen PH=Σ3

p. Similarly, let Π be an AND-compositional parameterized problem

whose unparameterized version Π is NP-complete. Then, if Π has a polynomialkernel the AND-Distillation Conjecture fails.

We can now formalize the discussion from the beginning of this section.

Theorem 5.1.7 ([20]) Longest Path does not admit a polynomial kernel un-less PH=Σ3

p.

Proof. The unparameterized version of Longest Path is known to be NP-complete [72]. We now give a composition algorithm for the problem. Given asequence (G1, k) . . . (Gt, k) of instances we output (G, k) where G is the disjointunion of G1 . . . Gt. Clearly G contains a path of length k if and only if Gi containsa path of length k for some i ≤ t. By Theorem 5.1.6 Longest Path does nothave a polynomial kernel unless PH=Σ3

p.

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An identical proof can be used to show that the Longest Cycle problemdoes not admit a polynomial kernel unless PH=Σ3

p. For many problems, it iseasy to give AND-composition algorithms. For instance, the “disjoint union”trick yields AND-composition algorithms for the Treewidth, Cutwidth andPathwidth problems, among many others. Coupled with Theorem 5.1.6 thisimplies that these problems do not admit polynomial kernels unless the AND-Distillation Conjecture fails. However, to this date, there is no strong complexitytheoretic evidence known to support the AND-Distillation Conjecture. Thereforeit would be interesting to see if such evidence could be provided.

5.2 Polynomial Parameter Transformations

For some problems, obtaining a composition algorithm directly is a difficult task.Instead, we can give a reduction from a problem that provably has no polynomialkernel unless PH=Σ3

p to the problem in question such that a polynomial kernelfor the problem considered would give a kernel for the problem we reduced from.We now define the notion of polynomial parameter transformations.

Definition 5.2.1 ([24]) Let P and Q be parameterized problems. We say thatP is polynomial parameter reducible to Q, written P ≤Ptp Q, if there exists apolynomial time computable function f : Σ∗ × N → Σ∗ × N and a polynomial p,such that for all (x, k) ∈ Σ∗ × N (a) (x, k) ∈ P if and only (x′, k′) = f(x, k) ∈ Qand (b) k′ ≤ p(k). The function f is called polynomial parameter transformation.

Proposition 5.2.2 ([24]) Let P and Q be the parameterized problems and Pand Q be the unparameterized versions of P and Q respectively. Suppose that Pis NP-complete and Q is in NP. Furthermore if there is a polynomial parametertransformation from P to Q, then if Q has a polynomial kernel then P also hasa polynomial kernel.

Proposition 5.2.2 shows how to use polynomial parameter transformations to showkernelization lower bounds. A notion similar to polynomial parameter transfor-mation was independently used by Fernau et al. [65] albeit without being explic-itly defined. We now give an example of how Proposition 5.2.2 can be usefulfor showing that a problem does not admit a polynomial kernel. In particular,we show that the Path Packing problem does not admit a polynomial kernelunless PH=Σ3

p. In this problem you are given a graph G together with an integerk and asked whether there exists a collection of k mutually vertex-disjoint pathsof length k in G. This problem is known to be fixed parameter tractable [11]and is easy to see that for this problem the “disjoint union” trick discussed inSection 5.1 does not directly apply. Thus we resort to polynomial parametertransformations.

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Theorem 5.2.3 Path Packing does not admit a polynomial kernel unless PH=Σ3p.

Proof. We give a polynomial parameter transformation from the Longest Pathproblem. Given an instance (G, k) to Longest Path we construct a graph G′

from G by adding k−1 vertex disjoint paths of length k. Now G contains a pathof length k if and only if G′ contains k paths of length k. This concludes theproof.

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Part II

New Methods

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Chapter 6

Algorithmic Method: ChromaticCoding

In this chapter we present a subexponential-time algorithm for the k-WeightedFeedback Arc Set in Tournaments problem, defined in the next section.Our algorithm is based on a novel version of the Color Coding technique initiatedin [11] and described in Section 2.7. The basic idea of Color Coding is to randomlycolor the input graph in order to make the solution more visible. That is, itis argued that with sufficiently high probability the random coloring unveils astructure that makes it easier to solve the problem instance. However, the morehelpful we require the coloring to be, the less likely it is that a random coloringwill be useful for our purposes. In the original Color Coding algorithm of Alon etal. [11] one required the coloring to color all vertices of the solution with differentcolors. Subsequently, the Divide and Color paradigm [95, 32] was introduced, andhere the coloring was required only to split the solution in half in a specific way,thus making it possible to speed up some of the algorithms of Alon et al. [11].

In the algorithm we present, the set of usable colorings is much larger. Weconsider the solution set as a k-edge graph and require the random coloring toproperly color this graph, that is the endpoints of every edge in this graph arecolored with different colors. We call this variation of the Color Coding techniqueChromatic Coding. Our algorithm runs in subexponential time, a trait uncommonto parameterized algorithms. In fact, to the author’s best knowledge the onlyparameterized problems for which non-trivial subexponential time algorithms areknown are bidimensional problems in planar graphs or graphs excluding a certainfixed graph H as a minor [43, 46, 49].

In order to derandomize our algorithm we construct a new kind of universalhash functions, that we coin universal coloring families. For integers m, k andr, a family F of functions from [m] to [r] is called a universal (m, k, r)-coloringfamily if for any graph G on the set of vertices [m] with at most k edges, thereexists an f ∈ F which is a proper vertex coloring of G. In the last section of the

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52

paper we give an explicit construction of a (10k2, k, O(√k))-coloring family F of

size |F| ≤ 2O(√

k) and an explicit universal (n, k, O(√k))-coloring family F of size

|F| ≤ 2O(√

k) log n. We believe that these constructions can turn out to be usefulto solve other edge subset problems in dense structures.

6.1 Tournament Feedback Arc Set

In a competition where everyone plays against everyone it is uncommon thatthe results are acyclic and hence one cannot rank the players by simply usinga topological ordering. A natural ranking is one that minimizes the number ofupsets, where an upset is a pair of players such that the lower ranked playerbeats the higher ranked one. The problem of finding such a ranking given thematch outcomes is the Feedback Arc Set problem restricted to tournaments.A tournament is a directed graph where every pair of vertices is connected byexactly one arc, and a feedback arc set is a set of arcs whose removal makesthe graph acyclic. Feedback arc sets in tournaments are well studied, both fromthe combinatorial [56, 64, 86, 89, 90, 124, 129, 130, 137], statistical [127] andalgorithmic [3, 6, 33, 91, 133, 134] points of view.

Unfortunately, the problem of finding a feedback arc set of minimum size inan unweighted tournament is NP-hard [6]. However, even the weighted versionof the problem admits a polynomial time approximation scheme [91] and hasbeen shown to be fixed parameter tractable [119]. One should note that theweighted generalization shown to admit a PTAS in [91] differs slightly from theone considered in this thesis. We consider the following problem:

k-Weighted Feedback Arc Set in Tournaments (k-FAST)Instance: A tournament T = (V,A), a weight function w : A →x ∈ R : x ≥ 1 and an integer k.Question: Is there an arc set S ⊆ A such that

∑e∈S w(e) ≤ k and

T \ S is acyclic?

The fastest previously known parameterized algorithm for k-FAST by Ramanand Saurabh [119] runs in time O(2.415k ·k4.752 +nO(1)), and it was an open prob-lem of Guo et al. [76] whether k-FAST can be solved in time 2k ·nO(1). We give a

randomized and a deterministic algorithm both running in time 2O(√

k log2 k)+nO(1).

6.1.1 Color and Conquer

Our algorithm consists of three steps. In the first step we reduce the instance to aproblem kernel with at most O(k2) vertices, showing how to efficiently reduce theinput tournament into one with O(k2) vertices, so that the original tournamenthas a feedback arc set of weight at most k, if and only if the new one has such a

53

1. Perform a data reduction to obtain a tournament T ′ of size O(k2).

2. Let t =√

8k. Color the vertices of T ′ uniformly at random with colors from1, . . . , t.

3. Let Ac be the set of arcs whose endpoints have different colors. Find a mini-mum weight feedback arc set contained in Ac, or conclude that no such feed-back arc set exists.

Figure 6.1: Outline of the algorithm for k-FAST.

set. In the second step we randomly color the vertices of our graph with t =√

8kcolors, and define the arc set Ac to be the set of arcs whose endpoints havedifferent colors. In the last step the algorithm checks whether there is a weight kfeedback arc set S ⊆ Ac. A summary of the algorithm is given in Figure 6.1.

6.1.2 Kernelization

For the first step of the algorithm we use the kernelization algorithm providedby Dom et al. [48]. They only show that the data reduction is feasible forthe unweighted case, while in fact, it works for the weighted case as well. Forcompleteness we provide a short proof of this. A triangle in what follows meansa directed cyclic triangle.

Lemma 6.1.1 k-FAST has a kernel with O(k2) vertices.

Proof. We give two simple reduction rules.

1. If an arc e is contained in at least k+1 triangles reverse the arc and reducek by w(e).

2. If a vertex v is not contained in any triangle, delete v from T .

The first rule is safe because any feedback arc set that does not contain thearc e must contain at least one arc from each of the k + 1 triangles containing eand thus must have weight at least k+1. The second rule is safe because the factthat v is not contained in any triangle implies that all arcs between N−(v) andN+(v) are oriented from N−(v) to N+(v). Hence for any feedback arc set S1 ofT [N−(v)] and feedback arc set S2 of T [N+(v)], S1 ∪S2 is a feedback arc set of T .

Finally we show that any reduced yes instance T has at most k(k+2) vertices.Let S be a feedback arc set of T with weight at most k. The set S contains at

54

most k arcs, and for every arc e ∈ S, aside from the two endpoints of e, there areat most k vertices that are contained in a triangle containing e, because otherwisethe first rule would have applied. Since every triangle in T contains an arc of Sand every vertex of T is in a triangle, T has at most k(k + 2) vertices.

6.1.3 Probability of a Good Coloring

We now proceed to analyze the second step of the algorithm. What we aim for,is to show that if T does have a feedback arc set S of weight at most k, then theprobability that S is a subset of Ac is at least 2−c

√k for some fixed constant c.

We show this by showing that if we randomly color the vertices of a k edge graphG with t =

√8k colors, then the probability that G has been properly colored is

at least 2−c√

k.

Lemma 6.1.2 If a graph on q edges is colored randomly with√

8q colors then

the probability that G is properly colored is at least (2e)−√

q/8.

Proof. Arrange the vertices of the graph by repeatedly removing a vertex oflowest degree. Let d1, d2, . . . , ds be the degrees of the vertices when they havebeen removed. Then for each i, di(s− i+1) ≤ 2q, since when vertex i is removedeach vertex had degree at least di. Furthermore, di ≤ s − i for all i, since thedegree of the vertex removed can not exceed the number of remaining vertices atthat point. Thus di ≤

√2q for all i. In the coloring, consider the colors of each

vertex one by one starting from the last one, that is vertex number s. When vertexnumber i is colored, the probability that it will be colored by a color that differsfrom all those of its di neighbors following it is at least (1 − di√

8q) ≥ (2e)−di/

√8q

because√

8q ≥ 2di. Hence the probability that G is properly colored is at least

s∏

i=1

(1 − di√8q

) ≥s∏

i=1

(2e)−di/√

8q = (2e)−√

q/8.

6.1.4 Solving a Colored Instance

Given a t-colored tournament T , we will say that an arc set F is colorful if noarc in F is monochromatic. An ordering σ of T is colorful if the feedback arcset corresponding to σ is colorful. An optimal colorful ordering of T is a colorfulordering of T with minimum cost among all colorful orderings. We now give analgorithm that takes a t-colored arc weighted tournament T as input and finds acolorful feedback arc set of minimum weight, or concludes that no such feedbackarc set exists.

55

Observation 6.1.3 Let T = (V1 ∪ V2 ∪ . . . ∪ Vt, A) be a t-colored tournament.There exists a colorful feedback arc set of T if and only if T [Vi] induces an acyclictournament for every i.

We say that a colored tournament T is feasible if T [Vi] induces an acyclic tour-nament for every i. Let ni = |Vi| for every i and let n be the vector [n1, n2 . . . nt].Let σ = v1v2 . . . vn be the ordering of V corresponding to a colorful feedbackarc set F of T . For every color class Vi of T , let v1

i v2i . . . v

ni

i be the order inwhich the vertices of Vi appear according to σ. Observe that since F is colorful,v1

i v2i . . . v

ni

i must be the unique topological ordering of T [Vi]. We exploit this togive a dynamic programming algorithm for the problem.

Lemma 6.1.4 Given a feasible t-colored tournament T , we can find a minimumweight colorful feedback arc set in O(t · nt+1) time and O(nt) space.

Proof. For an integer x ≥ 1, define Sx = v1 . . . , vx and Six = vi

1 . . . , vix. Let

S0 = Si0 = ∅. Notice that for any x there must be some x′ such that Sx∩Vi = Sx′.

Given an integer vector p of length t in which the ith entry is between 0 and ni,let T (p) be T [S1

p1∪ S2

p2. . . ∪ St

pt].

For a feasible t-colored tournament T , let Fas(T ) be the weight of the mini-mum weight colorful feedback arc set of T . Observe that if a t-colored tournamentT is feasible then so are all induced subtournaments of T , and hence the functionFas is well defined on all induced subtournaments of T . We proceed to provethat the following recurrence holds for Fas(T (p)).

Fas(T (p)) = mini : pi>0

(Fas(T (p − ei)) +∑

u∈V (T (p))

w∗(vipi

, u)) (6.1)

First we prove that the left hand side is at most the right hand side. Let i bethe integer that minimizes the right hand side. Taking the optimal ordering ofT (p− ei) and appending it with vi

pigives an ordering of T (p) with cost at most

Fas(T (p− ei)) +∑

u∈V (T (p))w∗(vi

pi, u).

To prove that the right hand side is at most the left hand side, take an optimalcolorful ordering σ of T (p) and let v be the last vertex of this ordering. There isan i such that v = vi

pi. Thus σ restricted to V (T (p − ei)) is a colorful ordering

of T (p− ei) and the total weight of the edges with startpoint in v and endpointin V (T (p− ei)) is exactly

∑u∈V (T (p)) w

∗(vipi, u). Thus the cost of σ is at least the

value of the right hand side of the inequality, completing the proof.Recurrence 6.1 naturally leads to a dynamic programming algorithm for the

problem. We build a table containing Fas(T (p)) for every p. There are O(nt)table entries, for each entry it takes O(nt) time to compute it giving the O(t·nt+1)time bound.

56

In fact, the algorithm provided in Lemma 6.1.4 can be made to run slightlyfaster by pre-computing the value of

∑u∈V (T (p)) w

∗(vipi, u)) for every p and i using

dynamic programming, and storing it in a table. This would let us reduce thetime to compute a table entry using Recurrence 6.1 from O(nt) to O(t) yieldingan algorithm that runs in time and space O(t · nt).

Lemma 6.1.5 k-FAST (for a tournament of size O(k2)) can be solved in ex-

pected time 2O(√

k log k) and 2O(√

k log k) space.

Proof. Our algorithm proceeds as described in Figure 6.1. The correctnessof the algorithm follows from Lemma 6.1.4. Combining Lemmata 6.1.1, 6.1.2,

6.1.4 yields an expected running time of O((2e)√

k/8) ·O(√

8k · (k2 + 2k)1+√

8k) ≤2O(

√k log k) for finding a feedback arc set of weight at most k if one exists. The

space required by the algorithm is O((k2 + 2k)1+√

8k) ≤ 2O(√

k log k).

The dynamic programming algorithm from Lemma 6.1.4 can be turned intoa divide and conquer algorithm that runs in polynomial space, at a small cost inthe running time.

Lemma 6.1.6 Given a feasible t-colored tournament T , we can find a minimumweight colorful feedback arc set in time O(n1+(t+2)·log n) in polynomial space.

Proof. By expanding Recurrence (6.1) ⌊n/2⌋ times and simplifying the righthand side we obtain the following recurrence.

Fas(T (p)) = minq≥0

q†·e=⌈n/2⌉

Fas(T (q)) + Fas(T \ V (T (q))) +∑

u∈V (T (q))v/∈V (T (q))

w∗(v, u) (6.2)

Recurrence 6.2 immediately yields a divide and conquer algorithm for theproblem. Let T (n) be the running time of the algorithm restricted to a sub-tournament of T with n vertices. For a particular vector q it takes at most n2

time to find the value of∑

u∈V (T (q)),v /∈V (T (q)) w∗(v, u). It follows that T (n) ≤

nt+2 · 2 · T (n/2) ≤ 2log n · n(t+2)·log n = n1+(t+2)·log n.

Theorem 6.1.7 k-FAST (for a tournament of size O(k2)) can be solved in ex-

pected time 2O(√

k log2 k) and polynomial space. Therefore, k-FAST for a tourna-ment of size n can be solved in expected time 2O(

√k log2 k) + nO(1) and polynomial

space.

57

6.2 Universal Coloring Families

For integers m, k and r, a family F of functions from [m] to [r] is called a universal(m, k, r)-coloring family if for any graph G on the set of vertices [m] with at mostk edges, there exists an f ∈ F which is a proper vertex coloring of G. An explicitconstruction of a (10k2, k, O(

√k))-coloring family can replace the randomized

coloring step in the algorithm for k-FAST. In this section, we provide such aconstruction.

Theorem 6.2.1 There exists an explicit universal (10k2, k, O(√k))-coloring fam-

ily F of size |F| ≤ 2O(√

k).

Proof. For simplicity we omit all floor and ceiling signs whenever these are notcrucial. We make no attempt to optimize the absolute constants in the O(

√k) or

in the O(√k) notation. Whenever this is needed, we assume that k is sufficiently

large.Let G be an explicit family of functions g from [10k2] to [

√k] so that every

coordinate of g is uniformly distributed in [√k], and every two coordinates are

pairwise independent. There are known constructions of such a family G with|G| ≤ kO(1). Indeed, each function g represents the values of 10k2 pairwise in-dependent random variables distributed uniformly in [

√k] in a point of a small

sample space supporting such variables; a construction is given, for example, in[7]. The family G is obtained from the family of all linear polynomials over afinite field with some kO(1) elements, as described in [7].

We can now describe the required family F . Each f ∈ F is described by asubset T ⊂ [10k2] of size |T | =

√k and by a function g ∈ G. For each i ∈ [10k2],

the value of f(i) is determined as follows. Suppose T = i1, i2, . . . , i√k, with

i1 < i2 < . . . < i√k. If i = ij ∈ T , define f(i) =√k + j. Otherwise, f(i) = g(i).

Note that the range of f is of size√k +

√k = 2

√k, and the size of F is at most

(10k2

√k

)|G| ≤

(10k2

√k

)kO(1) ≤ 2O(

√k log k) ≤ 2O(

√k).

To complete the proof we have to show that for every graph G on the set ofvertices [10k2] with at most k edges, there is an f ∈ F which is a proper vertexcoloring of G. Fix such a graph G.

The idea is to choose T and g in the definition of the function f that willprovide the required coloring for G as follows. The function g is chosen at randomin G, and is used to properly color all but at most

√k edges. The set T is chosen

to contain at least one endpoint of each of these edges, and the vertices in theset T will be re-colored by a unique color that is used only once by f . Usingthe properties of G we now observe that with positive probability the number ofedges of G which are monochromatic is bounded by

√k.

58

Claim 6.2.2 If the vertices of G are colored by a function g chosen at randomfrom G, then the expected number of monochromatic edges is

√k.

Proof. Fix an edge e in the graph G and j ∈ [√k]. As g maps the vertices

in a pairwise independent manner, the probability that both the end points ofe get mapped to j is precisely 1

(√

k)2. There are

√k possibilities for j and hence

the probability that e is monochromatic is given by√

k(√

k)2= 1√

k. Let X be the

random variable denoting the number of monochromatic edges. By linearity ofexpectation, the expected value of X is k · 1√

k=

√k.

Returning to the proof of the theorem, observe that by the above claim, withpositive probability, the number of monochromatic edges is upper bounded by√k. Fix a g ∈ G for which this holds and let T = i1, i2, . . . , i√k be a set of

√k

vertices containing at least one endpoint of each monochromatic edge. Considerthe function f defined by this T and g. As mentioned above f colors each of thevertices in T by a unique color, which is used only once by f , and hence we onlyneed to consider the coloring of G \ T . However all edges in G \ T are properlycolored by g and f coincides with g on G \ T . Hence f is a proper coloring of G,completing the proof of the theorem.

Remarks:

• Each universal (n, k, O(√k))-coloring family must also be an (n,

√k,O(

√k))-

hashing family, as it must contain, for every set S of√k vertices in [n], a

function that maps the elements of S in a one-to-one manner, since thesevertices may form a clique that has to be properly colored by a function ofthe family. Therefore, by the known bounds for families of hash functions(see, e.g., [117]), each such family must be of size at least 2Ω(

√k) log n.

Although the next result is not required for our results on the TournamentFeedback Arc Set problem, we present it here as it may be useful in similarapplications.

Theorem 6.2.3 For any n > 10k2 there exists an explicit universal (n, k, O(√k))-

coloring family F of size |F| ≤ 2O(√

k) log n.

Proof. Let F1 be an explicit (n, 2k, 10k2)-family of hash functions from [n] to10k2 of size |F1| ≤ kO(1) log n. This means that for every set S ⊂ [n] of size atmost 2k there is an f ∈ F1 mapping S in a one-to-one fashion. The existenceof such a family is well known, and follows, for example, from constructions ofsmall spaces supporting n nearly pairwise independent random variables takingvalues in [10k2]. Let F2 be an explicit universal (10k2, k, O(

√k))-coloring family,

as described in Theorem 6.2.1. The required family F is simply the family of all

59

compositions of a function from F2 followed by one from F1. It is easy to checkthat F satisfies the assertion of Theorem 6.2.3.

Finally, combining the algorithm from Theorem 6.1.7 with the universal col-oring family given by Theorem 6.2.1 yields a deterministic subexponential timepolynomial space algorithm for k-FAST.

Theorem 6.2.4 k-FAST can be solved in time 2O(√

k) + nO(1) and polynomialspace.

60

Chapter 7

Explicit Identification inW[1]-Hardness Reductions

In hardness reductions one has to code one combinatorial problem in the languageof another problem. To this end we require gadgets, small constructions that actas “subroutines” in our encoding. Typically one needs to construct gadgets thatencode selection of an element from a specific set, and gadgets that pass aroundinformation on which elements that have been selected. How this information isencoded greatly affects how the corresponding gadgets have to behave. In Sections7.1.1 and 7.1.2 we show that sometimes it is useful to assign identification numbersto the elements of the sets we are working with, because numbers often can beencoded in a combinatorial problem as a distribution of resources. In Section 7.2we show that the results from Sections 7.1.1 and 7.1.2 can be used to prove aninteresting trade-off between expressive power and computational tractability.

7.1 Hardness for Treewidth-Parameterizations

Due to Courcelle’s Theorem (Theorem 2.4.1) many problems admit FPT algo-rithms on graphs of bounded treewidth. It is natural to ask whether all problemsthat can be solved in polynomial time on graphs with constant treewidth ad-mit FPT algorithms when parameterized by the treewidth of the input graph.That is, are all problems that are solvable in time nf(tw(G)) also solvable intime g(tw(G))nO(1)? We show that the answer to this question is no, unlessFPT = W[1]. In particular, we show that the Equitable Coloring andCapacitated Dominating Set problems are W[1]-hard parameterized by thetreewidth of the input graph. These were the first problems to be shown W[1]-hard parameterized by the treewidth of the input graph.

61

62

7.1.1 Equitable Coloring

The notion of equitable coloring seems to have been first introduced by Meyerin 1973, where an application to scheduling garbage trucks is described [109].Recently, Bodlaender and Fomin have shown that determining whether a graphof treewidth at most t admits an equitable coloring, can be solved in time O(nO(t))[21].

We consider the parameterized complexity of Equitable Coloring (ECP)in graphs with bounded treewidth. We show that when ECP is parameterizedby (t, r), where t is the treewidth bound, and r is the number of color classes,the problem is W[1]-hard. We reduce from the Multicolor Clique problemdefined in Section 3.2. Our reduction is based on a methodology which is some-times termed as edge representation strategy. This strategy is very basic and isuseful for many reductions. Consider that the instance G = (V,E) of Multi-color Clique has its vertices colored by the integers 1, ..., k. Let V [i] denotethe set of vertices of color i, and let E[i, j], for 1 ≤ i < j ≤ k, denote the setof edges e = uv, where u ∈ V [i] and v ∈ V [j]. We also assume that |V [i]| = Nfor all i, and that |E[i, j]| = M for all i < j, that is, the vertex color classesof G, and also the edge sets between them, have uniform sizes. See Section 3.2for a justification of these assumptions. In this methodology our basic encodinggadgets correspond to edges which we call edge gadget. We generally have threekind of gadgets which are engineered together to get a reduction for the problem.

Selection Gadget: This gadget job is to select exactly one edge gadget amongedge gadgets corresponding to edges between any two color classes V [i] andV [j].

Coherence Gadget: This gadget makes sure that the edge gadgets selectedamong edge gadgets corresponding to edges emanating out from a particularcolor class V [i] has a vertex in common in V [i]. That is all the edgescorresponding to selected edge gadgets emanates from the same vertex inV [i].

Match Gadget: This gadget ensures that if we have selected an edge gadgetcorresponding to an edge (u, v) between V [i] and V [j] then the edge gadgetselected between V [j] and V [i] corresponds to (v, u).

In what follows next we show how to adhere to this strategy and form gadgets inthe context of reduction from Multicolor Clique to Equitable ColoringProblem in graphs with bounded treewidth. To show the desired reduction, weintroduce two more general problems. List analogues of equitable coloring havebeen previously studied by Kostochka, et al. [1].

List Equitable Coloring Problem (LECP): Given an inputgraph G = (V,E), lists Lv of colors for every vertex v ∈ V and a

63

positive integer r; does there exist a proper coloring f of G with rcolors that for every vertex v ∈ V uses a color from its list Lv suchthat for any two color class, Vi and Vj of the coloring f , ||Vi|−|Vj || ≤ 1?

Number List Coloring Problem (NLCP): Given an input graphG = (V,E), lists Lv of colors for every vertex v ∈ V , a functionh : ∪v∈V Lv → N, associating a number to each color, and a positiveinteger r; does there exist a proper coloring f of G with r colors thatfor every vertex v ∈ V uses a color from its list Lv, such that anycolor class Vc of the coloring f is of size h(c)?

Our main effort is in the reduction of the Multicolor Clique problem toNLCP. We will use the following sets of colors in our construction of an instanceof NLCP:

1. S = σ[i, j] : 1 ≤ i 6= j ≤ k2. S ′ = σ′[i, j] : 1 ≤ i 6= j ≤ k3. T = τi[r, s] : 1 ≤ i ≤ k, 1 ≤ r < s ≤ k, r 6= i, s 6= i4. T ′ = τ ′i [r, s] : 1 ≤ i ≤ k, 1 ≤ r < s ≤ k, r 6= i, s 6= i5. E = ǫ[i, j] : 1 ≤ i < j ≤ k6. E ′ = ǫ′[i, j] : 1 ≤ i < j ≤ k

Note that |S| = |S ′| = 2(

k2

), that is, there are distinct colors σ[2, 3] and σ[3, 2],

etc. In contrast, the colors τi[r, s] are only defined for r < s. We associate witheach vertex and edge of G a pair of (unique) identification numbers. The up-identification number v[up] for a vertex v should be in the range [n2 + 1, n2 + n],if G has n vertices and it could be chosen arbitrarily, but uniquely. Similarly, theup-identification number e[up] of an edge e of G can be assigned (arbitrarily, butuniquely) in the range [2n2 + 1, 2n2 +m], assuming G has m edges.

Choose a suitably large positive integer Z0, for example Z0 = n3, and definethe down-identification number v[down] for a vertex v to be Z0 − v[up], andsimilarly for the edges e of G, define the down-identification number e[down] tobe Z0 − e[up]. Choose a second large positive integer, Z1 >> Z0, for example, wemay take Z1 = n6.

Next we describe various gadgets and the way they are combined in the re-duction. First we describe the gadget which encodes the selection of the edgegoing between two particular color classes in G. In other words, we will think ofthe representation of a k-clique in G as involving the selection of edges (with eachedge selected twice, once in each direction) between the color classes of verticesin G, with gadgets for selection, and to check two things: (1) that the selections

64

in opposite color directions match, and (2) that the edges chosen from color classV [i] going to V [j] (for various j 6= i) all emanate from the same vertex in V [i].

There are 2(

k2

)groups of gadgets, one for each pair of color indices i 6= j. If

1 ≤ i < j ≤ k, then we will refer to the gadgets in the group G[i, j] as forwardgadgets, and we will refer to the gadgets in the group G[j, i] as backward gadgets.

If e ∈ E[i, j], then there is one forward gadget corresponding to e in thegroup G[i, j], and one backward gadget corresponding to e in the group G[j, i].The construction of these gadgets is described as follows.

The forward gadget corresponding to e = uv ∈ E[i, j].The gadget has a root vertex r[i, j, e], and consists of a tree of height 2. Thelist assigned to this root vertex contains two colors: σ[i, j] and σ′[i, j]. The rootvertex has Z1 + 1 children, and each of these is also assigned the two-elementlist containing the colors σ[i, j] and σ′[i, j]. One of the children vertices is distin-guished, and has 2(k − 1) groups of further children:

• e[up] children assigned the list σ′[i, j], ǫ[i, j].

• e[down] children assigned the list σ′[i, j], ǫ′[i, j].

• For each r in the range j < r ≤ k, u[up] children assigned the list σ′[i, j], τi[j, r].

• For each r in the range j < r ≤ k, u[down] children assigned σ′[i, j], τ ′i [j, r].

• For each r in the range 1 ≤ r < j, u[down] children assigned σ′[i, j], τi[r, j].

• For each r in the range 1 ≤ r < j, u[up] children assigned the list σ′[i, j], τ ′i [r, j].

The backward gadget corresponding to e = uv ∈ E[i, j].The gadget has a root vertex r[j, i, e], and consists of a tree of height 2. The listassigned to this root vertex contains two colors: σ[j, i] and σ′[j, i]. The root vertexhas Z1 + 1 children, and each of these is also assigned the two-element list con-taining the colors σ[j, i] and σ′[j, i]. One of the children vertices is distinguished,and has 2k groups of further children:

• e[up] children assigned the list σ′[j, i], ǫ′[i, j].

• e[down] children assigned the list σ′[j, i], ǫ[i, j].

• For each r in the range i < r ≤ k, v[up] children assigned the list σ′[j, i], τj [i, r].

• For each r in the range i < r ≤ k, v[down] children assigned σ′[j, i], τ ′j [i, r].

• For each r in the range 1 ≤ r < i, v[down] children assigned σ′[j, i], τj [r, i].

• For each r in the range 1 ≤ r < i, v[up] children assigned the list σ′[j, i], τ ′j [r, i].

65

The numerical targets (h function) .

1. For all c ∈ (T ∪ T ′), h(c) = Z0.

2. For all c ∈ (E ∪ E ′), h(c) = Z0.

3. For all c ∈ S, h(c) = (M − 1)(Z1 + 1) + 1.

4. For all c ∈ S ′, h(c) = (M − 1) + (Z1 + 1) + (k − 1)(M − 1)Z0.

That completes the formal description of the reduction from MulticolorClique to NLCP. We turn now to some motivating remarks about the design ofthe reduction.

Remarks on the colors, their numerical targets, and their role in thereduction.

(1). There are 2(

k2

)groups of gadgets. Each edge of G gives rise to two gadgets.

Between any two color classes of G there are precisely M edges, and thereforeM ·

(k2

)edges in G in total. Each group of gadgets therefore contains M gadgets.

The gadgets in each group have two “helper” colors. For example, the group ofgadgets G[4, 2] has the helper colors σ[4, 2] and σ′[4, 2]. The role of the gadgetsin this group is to indicate a choice of an edge going from a vertex in the colorclass V [4] of G to a vertex in the color class V [2] of G. The role of the 2

(k2

)

groups of gadgets is to represent the selection of(

k2

)edges of G that form a k-

clique, with each edge chosen twice, once in each direction. If i < j then thechoice is represented by the coloring of the gadgets in the group G[i, j], and theseare the forward gadgets of the edge choice. If j < i, then the gadgets in G[i, j]are backward gadgets (representing the edge selection in the opposite direction,relative to the ordering of the color classes of G). The numerical targets for thecolors in S ∪ S ′ are chosen to force exactly one edge to be selected (forward orbackward) by each group of gadgets, and to force the gadgets that are colored ina way that indicates the edge was not selected into being colored in a particularway (else the numerical targets cannot be attained). The numerical targets forthese colors are complicated, because of this role (which is asymmetric betweenthe pair of colors σ[i, j] and σ′[i, j]).

(2). The colors in T ∪T ′ and E∪E ′ are organized in symmetric pairs, and each pairis used to transmit (and check) information. Due to the enforcements alluded toabove, each “selection” coloring of a gadget (there will be only one possible in eachgroup of gadgets) will force some number of vertices to be colored with these pairsof colors, which can be thought of as an information transmission. For example,when a gadget in G[4, 2] is colored with a “selection” coloring, this indicates thatthe edge from which the gadget arises is selected as the edge from the color classV [4] of G, to the color class V [2]. There is a pair of colors that handles theinformation transmission concerning which edge is selected between the groups

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G[2, 4] and G[4, 2]. (Of course, something has to check that the edge selected inone direction, is the same as the edge selected in the other direction.) There issomething elegant about the dual-color transmission channel for this information.Each vertex and edge has two unique identification numbers, “up” and “down”,that sum to Z0. To continue the concrete example, G[4, 2] uses the (numberof vertices colored by the) pair of colors ǫ[2, 4] and ǫ′[2, 4] to communicate toG[2, 4] about the edge selected. The signal from one side consists of e[up] verticescolored ǫ[2, 4] and e[down] vertices colored ǫ′[2, 4]. The signal from the other sideconsists of e[down] vertices colored ǫ[2, 4] and e[up] vertices colored ǫ′[2, 4]. Thusthe numerical targets for these colors allow us to check whether the same edgehas been selected in each direction (if each color target of Z0 is met). Thereis the additional advantage that the amount of signal in each direction is thesame: in each direction a total of Z0 colored vertices, with the two paired colors,constitutes the signal. This means that, modulo the discussion in (1) above,when an edge is not selected, the corresponding non-selection coloring involvesuniformly the same number (i.e., Z0) of vertices colored “otherwise” for each ofthe (M − 1) gadgets colored in the non-selection way: this explains (part of) the(k − 1)(M − 1)Z0 term in (4) of the numerical targets.

(3). In a similar manner to the communication task discussed above, each of thek − 1 groups of gadgets G[i, ] need to check that each has selected an edge fromV [i] that originates at the same vertex in V [i]. Hence there are pairs of colorsthat provide a communication channel similar to that in (2) for this information.This role is played by the colors in T ∪ T ′. (Because of the bookkeeping issues,this becomes somewhat intricate in the formal definition of the reduction.)

The above remarks are intended to aid an intuitive understanding of thereduction. We now return to a more formal argument.

Claim 7.1.1 If G has a k-multicolor clique, then G′ is a yes-instance to NLCP.

Proof. The proof of this claim is relatively straightforward. The gadgets corre-sponding to the edges of a k-clique in G are colored in a manner that indicates“selected” (for both the forward and the backward gadgets) and all other gadgetsare colored in manner that indicates “non-selected”. The coloring that corre-sponds to “selected” colors the root vertex with the color σ[i, j], and this forcesthe rest of the coloring of the gadget. The coloring that corresponds to “non-selected” colors the root vertex with the color σ′[i, j]. In this case the coloringof the rest of the gadget is not entirely forced, but if the grandchildren verticesof the gadget are also colored with σ′[i, j], then all the numerical targets will bemet.

Claim 7.1.2 Suppose that Γ is a list coloring of G′ that meets all the numericaltargets. Then in each group of gadgets, exactly one gadget is colored in a waythat indicates “selection”.

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Proof. We argue this as follows. There cannot be two gadgets in any groupcolored in the “selection” manner, since this would make it impossible to meetthe numerical target for a color in S. If no gadget is colored in the “selection”manner, then again the targets cannot be met for the colors in S ∪S ′ used in thelists for this group of gadgets.

Claim 7.1.3 Suppose that Γ is a list coloring of G′ that meets all the numericaltargets. Then in each group of gadgets, every gadget that is not colored in a waythat indicates “selection” must have all of its grandchildren vertices colored withthe appropriate color in S ′.

Proof. This claim follows from Claim 7.1.2, noting that the numerical targetsfor the S ′ colors cannot be met unless this is so.

It follows from Claims 7.1.2 and 7.1.3, that if Γ is a list coloring of G′ thatmeets all the numerical targets, then in each group of gadgets, exactly one gadgetis colored in the “selection” manner, and all other gadgets are colored in a com-pletely determined “nonselection” manner. Each “selection” coloring of a gadgetproduces a numerical signal (based on vertex and edge identification numbers)carried by the colors in T ∪ T ′ and E ∪ E ′, with two signals per color. The targetof Z0 for these colors can only be achieved if the selection colorings indicate aclique in G.

Theorem 7.1.4 NLCP is W[1]-hard for trees, parameterized by the number ofcolors that appear on the lists.

The reduction from NLCP to LECP is almost trivial, achieved by paddingwith isolated vertices having single-color lists. The reduction from LECP to ECPis described as follows. Create a clique of size r, the number of colors occurring onthe lists, and connect the vertices of this clique to the vertices of G′ in a mannerthat enforces the lists. Since G′ is a tree, the treewidth of the resulting graph isat most r. We have:

Theorem 7.1.5 Equitable Coloring is W[1]-hard parameterized by treewidthand the number r of colors.

7.1.2 Capacitated Dominating Set

In this section we show that Capacitated Dominating Set is W [1]-hard whenparameterized by treewidth and solution size. Just as in the previous section, wereduce from Multicolor Clique. See Section 3.2 for the definition of thisproblem. In fact, we will reduce to a slightly modified version of CapacitatedDominating Set, Marked Capacitated Dominating Set where we mark

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some vertices and demand that all marked vertices must be in the dominating set.We can then reduce from Marked Capacitated Dominating Set to Capac-itated Dominating Set by attaching k + 1 leaves to each marked vertex andincreasing the capacity of each marked vertex by k+ 1. It is easy to see that thenew instance has a k-capacitated dominating set if and only if the original onehad a k-capacitated dominating set that contained all marked vertices, and thatthis operation does not increase the treewidth of the graph. Thus, to prove thatCapacitated Dominating Set is W [1]-hard when parameterized by treewidthand solution size, it is sufficient to prove that Marked Capacitated Dominat-ing Set is. We will show how given an instance (G, k) of Multicolor Clique,we can build an instance (H, c, k′) of Marked Capacitated Dominating Setsuch that

• k′ = 7k(k − 1) + 2k,

• G has a clique of size k if and only if H has a capacitated dominating setof size k′, and

• the treewidth of H is O(k4).

We employ a strategy similar to the edge representation strategy we usedto show that Equitable Coloring is W[1]-hard parameterized by treewidth.The main difference in our approach is that the coherence gadget is switchedout with a vertex selection gadget and the vertex and edge selection gadgets arejoined together in such a way that one only can pick an edge if one also picks itsendpoints.

For a pair of distinct integers i, j, let E[i, j] be the set of edges with oneendpoint in V [i] and the other in V [j]. Without loss of generality, we will assumethat |V [i]| = N and |E[i, j]| = M for all i, j, i 6= j. To each vertex v weassign a unique identification number vup between N + 1 and 2N , and we setvdown = 2N − vup. For two vertices u and v, by adding an (A,B)-arrow from uto v we will mean adding A subdivided edges between u and v and attaching Bleaves to v (see Fig. 7.1). Now we describe how to build the graph H for a giveninstance (G = (V [1] ∪ V [2] · · · ∪ V [k], E), k) of Multicolor Clique.

(A,B)u v

=vu

12

A

2 B1

Figure 7.1: Adding an (A,B)-arrow from u to v.

For every integer i between 1 and k we add a marked vertex xi that has aneighbor v for every vertex v in V [i]. For every j 6= i, we add a marked vertex yij

69

and a marked vertex zij . Now, for every vertex v ∈ V [i] and every integer j 6= iwe add a (vup, vdown)-arrow from v to yij and a (vdown, vup)-arrow from v to zij .Finally we add a set Si of k′ + 1 vertices and make every vertex in Si adjacent toevery vertex v with v ∈ V [i]. See Fig. 7.2 for an illustration.

......

...

...v (vup , v

down)

(vdown, vup)(vup

, v down)(v down, v up

)(v up, v dow

n)

(v down, v u

p)

x2

y21

z21

y23

z23

y2k

z2k

S2

qji

pji

qij

pij

.........

xij

e (udown, uup)

(uup, udown)

(v up, v dow

n)

(v down, v up

)

Sij

(e = u, v)

Figure 7.2: Vertex and edge selection gadgets.

Similarly, for every pair of integers i, j with i < j, we add a marked vertex xij

with a neighbor e for every edge e in E[i, j]. Moreover, we add four new markedvertices pij, pji, qij, and qji. For every edge e = u, v in E[i, j] with u ∈ V [i] andv ∈ V [j], we add a (udown, uup)-arrow from e to pij, a (uup, udown)-arrow from e toqij , a (vdown, vup)-arrow from e to pji and a (vup, vdown)-arrow from e to pji. Wealso add a set Sij of k′ +1 vertices and make every vertex in Sij adjacent to everyvertex e with e ∈ E[i, j]. See Fig. 7.2 for an illustration.

Finally, we add a marked vertex rij and a marked vertex sij for every i 6= j.For every i 6= j, we add (2N, 0)-arrows from yij to rij, from pij to rij , from zij

to sij , and from qij to sij (see Fig. 7.3). This concludes the description of thegraph H .

Figure 7.3: Vertex-Edge incidence gadget

We now describe the capacities of the vertices. For every i 6= j, the vertex xi

has capacity N − 1, the vertex xij has capacity M − 1, the vertices yij and zij

both have capacity 2N2, the vertices pij and qij have capacity 2NM , and both rij

and sij have capacity 2N . For all other vertices, their capacity is equal to theirdegree in H .

Observation 7.1.6 The treewidth of H is O(k4).

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Proof. If we remove all marked vertices (⋃k

i=1 Si and⋃

i6=j Sij), a total of O(k4)vertices, from H , we obtain a forest. As deleting a vertex reduces the treewidthby at most one, this concludes the proof.

Lemma 7.1.7 If G has a multicolor clique C = v1, v2, . . . , vk then H has acapacitated dominating set D of size k′ containing all marked vertices.

Proof. For every i < j let eij be the edge from vi to vj in G. In additionto all the marked vertices, let D contain vi and eij for every i < j. Clearly Dcontains exactly k′ vertices, so it remains to prove that D is indeed a capacitateddominating set.For every i < j, let xi and xij dominate all their neighbors exceptfor vi and eij respectively. The vertices vi and eij can dominate all their neighbors,since their capacity is equal to their degree. Let rij dominate vdown

i of the verticesin the (2N, 0)-arrow from yij, and vup

i of the vertices of the (2N, 0)-arrow frompij. Similarly let sij dominate vup

i of the vertices in the (2N, 0)-arrow from zij ,and vdown

i of the vertices of the (2N, 0)-arrow from qij . Finally, for every i 6= j welet yij, zij , pij and qij dominate all their neighbors that have not been dominatedyet. One can easily check that every vertex of H will either be a dominator ordominated in this manner, and that no dominator dominates more vertices thanits capacity.

Lemma 7.1.8 If H has a capacitated dominating set D of size k′ containing allmarked vertices, then G has a multicolor clique of size k.

Proof. Observe that for every integer 1 ≤ i ≤ k, there must be a vi ∈ V [i] suchthat vi ∈ D. Otherwise we have that Si ⊂ D and, since |Si| > k′, we obtain acontradiction. Similarly, for every pair of integers i, j with i < j there must bean edge eij ∈ E[i, j] such that eij ∈ D. We let eji = eij . Since |D| ≤ k′ it followsthat these are the only unmarked vertices in D. Since all the unmarked verticesin D have capacity equal to their degree, we can assume that each such vertexdominates all its neighbors. We now proceed with proving that for every pair ofintegers i,j with i 6= j, eij = uv is incident to vi. We prove this by showing thatif u ∈ V [i] then vup

i + udown = 2N .Suppose for a contradiction that vup

i + udown < 2N . Observe that each vertexof T = (N(yij)∪N(rij)∪N(pij)) \ (N(vi)∪N(eij)) must be dominated by eitheryij, rij, or pij . However, by our assumption that vup

i +udown < 2N , it follows that|T | = 2N2 + 4N + 2MN − (vup

i + udown) > 2N2 + 2N + 2MN . The sum of thecapacities of yij, rij, and pij is exactly 2N2 + 2N + 2MN . Thus it is impossiblethat every vertex of T is dominated by one of yij, rij , and pij, a contradiction. Ifvup

i +udown > 2N then vdowni +uup < 2N , and we can apply an identical argument

for zij, sij , and qij .

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Thus, it follows that for every i 6= j there is an edge eij incident both to vi

and to vj . Thus v1, v2, . . . , vk forms a clique in G. As any k-clique in G is amulticolor clique this completes the proof.

Observation 7.1.6, Lemma 7.1.7 and Lemma 7.1.8 immediately imply the fol-lowing theorem.

Theorem 7.1.9 Capacitated Dominating Set parameterized by treewidthand solution size is W [1]-hard.

7.2 Hardness for Cliquewidth-Parameterizations

By the seminal result of Courcelle [35] (Theorem 2.4.1), all problems expressiblein MSO2 logic are FPT when parameterized by the treewidth of the input graph.The result of Courcelle was extended by Courcelle, Makowsky and Rotics [36](Theorem 2.4.4) who proved that all problems expressible in MSO1 logic areFPT in parameterized by the cliquewidth of the input graph. For several of theproblems expressible in MSO2 but not MSO1, like Hamiltonian Cycle andEdge Dominating Set algorithms with running times on the form nf(cwd(G))

were derived, but no FPT algorithms for Hamiltonian Cycle and Edge Dom-inating Set were found. Another classical problem that is known to be FPTparameterized by treewidth and solvable in nf(cwd(G)) time is Graph Color-ing. The question on the existence of fixed parameter tractable algorithms (withclique-width being the parameter) for all these problems (or their generalizations)was asked by Gerber and Kobler [73], Kobler and Rotics [96, 97], Makowsky,Rotics, Averbouch, Kotek, and Godlin [105, 74]. In the next sections we showthat Graph Coloring, Hamiltonian Cycle and Edge Dominating Setall are W[1]-hard parameterized by the cliquewidth of the input graph. This im-plies that unless FPT = W[1], any extension of Courcelle’s theorem to graphs ofbounded cliquewidth can not apply to all problems expressible in MSO2.

7.2.1 Graph Coloring — Chromatic Number

In this section, we prove that Graph Coloring is W [1]-hard parameterized byclique-width.

Graph Coloring (or Chromatic Number): The chromatic num-ber of a graph G = (V (G), E(G)) is the smallest number of colorsχ(G) needed to color the vertices of G so that no two adjacent ver-tices are of the same color.

Our reduction is from the Equitable Coloring problem parameterizedby the number r of colors used, and the treewidth of the input graph. In the

72

Equitable Coloring problem one is given a graph G and integer r and askedwhether G can be properly r-colored in such a way that the number of verticesin any two color classes differs by at most 1. Notice that if n is divisible by rthis implies that all color classes must contain the same number of vertices. Inour reduction we will assume that in the instance we reduce from, n is divisibleby r. For a justification of this assumption, if r does not divide n we can add aclique of size n+ r−⌊n

r⌋r to G. We reduce from the exact version of Equitable

Coloring, that is, the version where we are looking for an equitable coloring ofG with exactly r colors. Recall that in Section 7.1.1 we proved that EquitableColoring is W [1]-hard parameterized by the treewidth t of the input graph andthe number of colors r.

Construction: On input (G, r) to Equitable Coloring, we construct aninstance (G′, r′) of Graph Coloring as follows. We start with a copy of G andlet r′ = r+nr. We now add a clique P of size r′ to G′. The clique P will functionas a palette in our reduction, as we have to use all r′ available colors to properlycolor it. We partition P into r + 1 parts as follows, P = PM ∪ P1 ∪ P2 · · · ∪ Pr

where PM has size r and Pi has size n for every i. We call PM the main palette,and denote the vertices in PM by pi for 1 ≤ i ≤ r. We add edges between everyvertex of P \ PM and every vertex of the copy of G. For each vertex u ∈ V (G)we assign a vertex uPi

∈ Pi for every i. Now, for every 1 ≤ i ≤ r we add a setSi of vertices. For each vertex u ∈ V (G) we make a vertex uSi

in Si for every1 ≤ i ≤ r, and make uSi

adjacent to u and the entire palette P except for uPi

and pi. We conclude the construction by adding a clique Ci of n r−1r

verticesand making every vertex of Ci adjacent to all of the vertices of Si and the entirepalette except for Pi.

Lemma 7.2.1 If G has an equitable r-coloring ψ, then G′ has an r′-coloring φ.

Proof. We construct a coloring φ of G′ as follows. The coloring φ colors the copyof G in G′ in the same way that ψ colors G. We color the palette, assigning aunique color to each vertex and making sure that the main palette PM is coloredusing the same colors that are used to color the vertices of G. For every vertexuSi

we color uSiwith φ(pi) if φ(u) 6= φ(pi) and with φ(uPi

) if φ(u) = φ(pi). Wecolor every vertex of Ci with some color from Pi (a color used to color a vertex ofPi). To do this we need n r−1

rdifferent colors from Pi. Since exactly n/r vertices

of G are colored with φ(pi), exactly n r−1r

of Si are colored with φ(pi) and thusn/r vertices of Si are colored with colors of Pi. Hence there are n r−1

rcolors of Pi

available to color Ci. Thus, φ is a proper r′-coloring of G concluding the proof.

Lemma 7.2.2 If G′ has an r′-coloring φ, then G has an equitable r-coloring ψ.

73

Proof. We prove that the restriction of φ to the copy of G in G′ in fact is anequitable r-coloring of G. Since φ can only use the colors of PM , φ is a properr-coloring of G. It remains to prove that for any i between 1 and r, at mostn/r vertices of G are colored with φ(pi). Suppose for contradiction that there isan i such that more than n/r vertices of G are colored with φ(pi). Then thereare more than n/r vertices of Si that are colored with colors of Pi. Since eachsuch vertex must take a different color from Pi, there are less than n r−1

rdifferent

colors of Pi available to color the vertices of Ci. However, since Ci is a clique onn r−1

rvertices that must be colored with colors of Pi, this is a contradiction.

Lemma 7.2.3 If the treewidth of G is t, then the cliquewidth of G′ is at mostk = 3 · 2t−1 + 7r + 3. Furthermore, an expression tree of width k for G′ can becomputed in FPT time.

Proof. By Theorem 1.2.1, we can compute an expression tree for G of width atmost 3·2t−1 in FPT time. Our strategy is as follows. We first show how to modifythe expression tree to give a width k expression tree for G′ \ (PM

⋃ri=1Ci). Then

we change this tree into an expression tree for G′. In order to give an expressiontree for G′ we introduce the following extra labels.

• For every 1 ≤ i ≤ r the labels αi, αLi and αR

i for vertices in Pi.

• For every 1 ≤ i ≤ r the labels βi, βLi and βR

i for vertices in Si.

• For every 1 ≤ i ≤ r the label ζi for vertices in Ci.

• A “work” label – γW , and a label γM for PM .

In the expression tree for G, we replace every introduce-node i(v) with a smallexpression tree Ti(v). In Ti(v), the vertex v is introduced with label γW and thevertices vP1 , . . . , vPr

and vS1 , . . . , vSrare introduced with labels α1, . . . , αr and

β1, . . . , βr respectively. Also, γW is joined to β1, . . . , βr and for every p, βp isjoined with every label in αq : q 6= p. Also, for every p 6= q, αp is joined withαq. Finally, γW is relabelled to i.

Now, for every union node in the expression tree (not the union nodes insidethe Ti’s) we add extra vertices on the edges incident to this node. On the edgefrom the node to its left child, we add nodes that relabel αp to αL

p and βp to βLp

for every p. Similarly, on the edge from the union node to its right child, we addnodes that relabel αp to αR

p and βp to βRp for every p. Finally, on the edge from

the union node to its parent we add nodes that first join every αLp with every βR

q

and αRq , join every αR

p with every βLq , and then relabel every αL

p and αRp to αp

and every βLp and βR

p to βp.

74

To conclude the construction of G′ \ (PM⋃r

i=1Ci) we need to add some extranodes above the root of the expression tree. We add the edges between P \ PM

and G by joining every αp with all labels used for constructing G.

We now need to add the construction of PM and⋃r

i=1Ci to our expressiontree. We start by making Cp for every p between 1 and r. For every p we adda clique on n r−1

rvertices labelled ζp. Every ζp is joined to βp and for every pair

p 6= q, ζp is joined with αq.

Finally, we add the construction of PM . For every i, we introduce the vertexpi with label γW , join γW to αj and ζj for every j, γW with βj for every j 6= i andfinally join γW to γM and relabel γW to γM . This concludes the construction ofG′. Notice that this expression tree for G′ uses k = 3 · 2t−1 + 9r + 3 labels.

Lemmas 7.2.1, 7.2.2 and 7.2.3 together imply the following result.

Theorem 7.2.4 The Graph Coloring problem is W [1]-hard when parame-terized by clique-width. Moreover, this problem remains W [1]-hard even if theexpression tree is given.

7.2.2 Edge Dominating Set

In this section, we show that Edge Dominating Set is W [1]-hard parameter-ized by clique-width.

Edge Dominating Set: Given a graph G = (V,E), find a minimumset of edges X ⊆ E(G) such that every edge of G is either includedin X, or it is adjacent to at least one edge of X. The set X is calledan edge dominating set of G

Our reduction is from a variant of Capacitated Dominating Set problem,Exact Saturated Capacitated Dominating Set: A capacitated graph isa pair (G, c), where G is a graph and c : V (G) → N is a capacity function suchthat 1 ≤ c(v) ≤ deg(v) for every vertex v ∈ V (G) (sometimes we simply saythat G is a capacitated graph if the capacity function is clear from the context).A set S ⊆ V (G) is called a capacitated dominating set if there is a dominationmapping f : V (G) \ S → S which maps every vertex in (V (G) \ S) to one ofits neighbors such that the total number of vertices mapped by f to any vertexv ∈ S does not exceed its capacity c(v). We say that for a vertex u ∈ S, verticesin the set f−1(u) are dominated by u. The Capacitated Dominating Setproblem is formulated as follows: given a capacitated graph (G, c) and a positiveinteger k, determine whether there exists a capacitated dominating set S for Gcontaining at most k vertices. Recall that in Section 7.1.2 we proved this problemis W [1]-hard when parameterized by treewidth.

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For the intractability proof of Edge Dominating Set, we need a specialvariant of Capacitated Dominating Set problem which we call Exact Sat-urated Capacitated Dominating Set. Given a capacitated dominating setS, a vertex v ∈ S is called saturated if the corresponding domination mapping fmaps c(v) vertices to v, that is, |f−1(v)| = c(v). A capacitated dominating setS ⊆ V (G) is called saturated if there is a domination mapping f which saturatesall vertices of S. In Exact Saturated Capacitated Dominating Set a ca-pacitated graph (G, c) and a positive integer k is given. The question is whetherG has a saturated capacitated dominating set S with exactly k vertices.

Lemma 7.2.5 The Exact Saturated Capacitated Dominating Set prob-lem is W [1]-hard when parameterized by clique-width. Moreover, this problemremains W [1]-hard even if the expression tree is given.

Proof. We show a stronger statement, namely that the Exact SaturatedCapacitated Dominating Set problem is W [1]-hard when parameterized bythe treewidth of the input graph. The statement of the lemma then follows di-rectly from Theorem 1.2.1. To show that the Exact Saturated CapacitatedDominating Set problem is W [1]-hard when parameterized by the treewidthof the input graph, we consider the reduction from Multicolor Clique toCapacitated Dominating Set presented in Section 7.1.2. We tweak the re-duction a little bit by for every i, j ≤ k giving all the neighbours of xi and xij

capacity one less than their degree instead of their degree. Let c′ be the tweakedcapacity function. One can now verify that (H, c′, k′) has a capacitated dominat-ing set of size at most k′ if and only if (H, c, k′) does, and that if (H, c′, k′) has acapacitated dominating set of size at most k′ then this set has size exactly k′ andis saturated. Observe also that the reduction from the marked version of Ca-pacitated Dominating Set to the original version preserves exact saturatingcapacitated dominating sets.

We now proceed to show that Edge Dominating Set is W [1]-hard param-eterized by clique-width by giving a reduction from Exact Saturated Domi-nating Set. We start with descriptions of auxiliary gadgets.

Auxiliary gadgets: Let s ≤ t be positive integers. We construct a graph Fs,t

with the vertex set x1, . . . , xs, y1, . . . , ys, z1, . . . , zt and edges xiyi, 1 ≤ i ≤ s andyizj, 1 ≤ i ≤ s and 1 ≤ j ≤ t. Basically we have complete bipartite graph betweenyi’s and zj ’s with pendent vertices attached to yi’s. The vertices z1, z2, . . . , zt arecalled roots of Fs,t.

Graph Fs,t has the following property.

Lemma 7.2.6 Any set of s edges incident to vertices y1, . . . , ys forms an edgedominating set in Fs,t. Furthermore, let G be a graph obtained by the union of

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Uiai

bi

wi

X

Ri

vi vj

Uj

Figure 7.4: Graph G′

Fs,t with some other graph H such that V (Fs,t)∩V (H) = z1, . . . , zt. Then everyedge dominating set of G contains at least s edges from Fs,t.

The proof of the lemma follows from the fact that every edge dominating setincludes at least one edge from E(yi) for i ∈ 1, . . . , s.

Now we describe our reduction. Let (G, c) be a capacitated graph with thevertex set u1, . . . , un, and k be a positive integer. For every vertex ui, theset Ui with c(ui) vertices is introduced, and then vertex sets v1, . . . , vn andw1, . . . , wn are added. For every edge uiuj ∈ E(G), all vertices of Ui are joinedwith vj and all vertices of Uj are joined with vi by edges. Then every vertex vi isjoined to its counterpart wi and to every vertex vi we add one additional leaf (apendent vertex). Now vertex sets a1, . . . , an and b1, . . . , bn are constructed,and vertices ai are made adjacent to all vertices of Ui, wi and bi. For every vertexbi, a set Ri of c(ui)+1 vertices is added and bi is made adjacent to all the verticesin Ri. Then we add to every vertex of R1 ∪ R2 ∪ · · · ∪ Rn a path of length two.Let X be the set of middle vertices of these paths. We denote the obtained graphby G′ (see Fig 7.4). Finally, we introduce three copies of Fs,t:

• a copy of Fn−k,n with roots a1, . . . , an,

• a copy of Fk,n with roots b1, . . . , bn, and a

• a copy of Fn,r where r =n∑

i=1

c(ui) with roots in X.

Let this final resulting graph be H .

Lemma 7.2.7 A graph G has an exact saturated dominating set of the size k ifand only if H has an edge dominating set of cardinality at most 2n+ r.

Proof. Let S be an exact saturated dominating set of the size k in G and fbe its corresponding domination mapping. For convenience (without loss of agenerality) we assume that S = u1, . . . , uk. We construct the edge dominatingset as follows. First we select an edge emanating from every vertex in the setv1, . . . , vn. For every vertex vi, 1 ≤ i ≤ k, the edge viwi is selected. Now let us

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assume that k < i ≤ n and f(ui) = uj. We choose a vertex u in Uj which is notincident to already chosen edges and add the edge uvi to our set. Notice that wealways have such a choice of u ∈ Uj as c(uj) = |Uj |. We observe that these edgesalready dominates all the edges in the sets E(vi), 1 ≤ i ≤ n, and in sets E(u) foru ∈ U1 ∪ · · · ∪Uk ∪ w1, . . . , wk. Now we add n− k edges from Fn−k,n which areincident to vertices in ak+1, . . . , an and k edges from Fk,n which are incidentto b1, . . . , bk. Then r − n matching edges joining vertices of Rk+1, . . . , Rn tothe vertices of X are included in the set. Finally, we add n edges form Fn,r

which are incident to vertices of X which are adjacent to vertices of R1, . . . , Rk.Since S is an exact capacitated dominating set,

∑ki=1(c(ui) + 1) = n, and from

our description it is clear that the resulting set is an edge dominating set of size2n+ r for H .

We proceed by proving the other direction of the equivalence. Let L be anedge dominating set of cardinality at most 2n+ r. The set L is forced to containat least one edge from every E(vi), at least n − k edges from Fn−k,n, at leastk edges from Fk,n, and at least one edge from E(x) for all x ∈ X because ofpendent edges. This implies that |L| = 2n + r, and L contains exactly one edgefrom every E(vi), exactly n−k edges from Fn−k,n, exactly k edges from Fk,n, andexactly one edge from E(x) for all x ∈ X. Every edge aibi needs to be dominatedby some edge of L, in particular it must be dominated from either an edge ofFn−k,n, or Fk,n. Let I = i : ai is incident to an edge from L ∩ E(Fn−k,n) andJ = j : bj is incident to an edge from L ∩ E(Fk,n). The above constraints onthe set L implies that |I| = n − k, |J | = k, and these sets form a partition of1, . . . , n. The edges which join vertices bi and Ri for i ∈ I are not dominated byedges from L∩E(Fk,n). Hence to dominate these edges we need at least

∑i∈I |Ri|

edges which connect sets Ri andX. Since at least n edges of Fn,r are included in L,we have that

∑i∈I |Ri| ≤ r−n and

∑j∈J |Rj | = r−∑i∈I |Ri| ≥ r− (r− n) ≥ n.

Let S = uj : j ∈ J. Clearly, |S| = k. Now we show that S is a saturatedcapacitated dominating set. For j ∈ J , edges which join a vertex aj to Uj andwj are not dominated by edges from L ∩ E(Fn−k,k), and hence they have to bedominated by edges from sets E(vi). Since n ≤∑j∈J |Rj| =

∑j∈J(|Uj|+1), there

are exactly n such edges, and every such edge must be dominated by exactly oneedge from L. An edge ajwj can only be dominated by edge vjwj . We also knowthat L ∩ E(vi) 6= ∅ for all i ∈ 1, . . . , n and hence for every vi, i /∈ J , thereis exactly one edge which joins it with some vertex u ∈ Uj for some j ∈ J .Furthermore, all these edges are not adjacent, that is, they form a matching.We define f(ui) = uj for i /∈ J . From our construction it follows that f is adomination mapping for S and S is an exact saturated dominating set in G.

The next lemma shows that if the graph G we started with has boundedclique-width then H also has bounded clique-width.

Lemma 7.2.8 If cwd(G) ≤ t then cwd(H) ≤ 2t + 16, and an expression tree

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for H of width at most w = 2t+ 16 can be constructed in a polynomial time fromthe expression tree for G.

Proof. The graph G is of clique-width at most t. Suppose that the expressiontree for G uses t-labels α1, . . . , αt. To construct the expression tree for H weneed following additional labels:

• Labels β1, . . . , βt for the vertices in U1, . . . , Un.

• Labels ξ1, ξ2, and ξ3 for attaching Fn−k,n, Fk,n and Fn,r respectively.

• Labels ζ1, . . . , ζ4 for marking some vertices like w1, . . . , wn.

• Working labels γ1, . . . , γ9.

When a vertex ui ∈ V (G) labeled αj is introduced, we perform following set ofoperations. First we introduce following vertices with some working labels: vi

with label γ1, c(ui) vertices of Ui with label γ2, the vertex wi with label γ3, andthe additional vertex (the leaf attached to vi) with label γ4. Now we join thevertex labelled with γ1 to vertices labelled with γ3 and γ4 (basically joining vi

with wi and its pendent leaf). Finally, we relabel γ4 to ζ1 and γ1 to βj. Now weintroduce vertices ai and bi with labels γ5 and γ6 respectively. Then we join thevertex labelled γ4 (ai) with all the vertices labelled with γ2, γ3 and γ6 (Ui, wi, bi).The join operation is followed by relabeling γ3 to ζ2, γ2 to αj and γ5 with ξ1.

Now we want to make the vertices of Ri and the paths attached to it. Todo so we perform following operations c(ui) + 1 times: (a) introduce three nodeslabelled with γ7, γ8 and γ9 (b) join γ6 with γ7, γ7 with γ8 and γ8 with γ9 and(c) finally we relabel γ6 to ξ2, γ7 to ζ3, γ8 to ξ3 and γ9 to ζ4. We omit the unionoperations from the description and assume that if some vertex is introduced thenthis operation is performed.

If in the expression tree of G, we have join operation between two labels sayαi and αj then we simulate this by applying join operations between αi and βj

and αj and βi. The relabel operation in the expression tree of G, that is, relabelαi to αj is replaced by relabel αi to αj and relabel βi to βj. Union operations inthe expression tree is done as before.

Finally to complete the expression tree for H , we need to add Fn−k,n, Fk,n andFn,r. Notice that all the vertices in a1, . . . , an, b1, . . . , bn and X are labelledξ1, ξ2 and ξ3 respectively. From here we can easily add Fn−k,n, Fk,n and Fn,r

with root vertices a1, . . . , an, b1, . . . , bn and X respectively by using workinglabels. This concludes the description for the expression tree for H .

Lemmas 7.2.7 and 7.2.8 together imply the following result.

Theorem 7.2.9 The Edge Dominating Set problem is W [1]-hard when pa-rameterized by clique-width. Moreover, the problem remains W [1]-hard even ifthe expression tree is given.

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7.2.3 Hamiltonian Cycle

In this section we show that the Hamiltonian Cycle problem is W [1]-hard forthe graphs of bounded clique-width.

Hamiltonian Cycle: Given a graph G, check whether there existsa cycle passing through every vertex of G.

Our reduction is from the Capacitated Dominating Set problem describedin Section 7.1.2 and shown to be W [1]-hard in Theorem 7.1.9. We describe a fewauxiliary gadgets.

Auxiliary gadgets: We denote by L1, the graph with the vertex set x, y, z, a, b, c, dand the edge set xa, ab, bc, cd, dy, bz, cz. Let P1 be the path xabzcdy, andP2 = xabcdy. (See Fig. 7.5.)

We abstract a property of this graph in the following lemma.

Lemma 7.2.10 Let G be a Hamiltonian graph such that G[V ′] is isomorphic toL1. Furthermore, if all edges in E(G) \ E(G[V ′]) incident to V ′ are incident tothe copies of the vertices x, y, and z in V ′, then every Hamiltonian cycle in Geither includes the path P1, or the path P2 as a segment.

Our second auxiliary gadget is the graph L2. This graph has x, y, z, s, t, a,b, c, d, e, f, g, h as its vertex set. We first include following xa, ab, bz, cz, cd, dy,se, ef , fb, ch, hg, gt in its edge set. Then x, y-path of length 10 xw1 · · ·w9y isadded, and edges fw3, w1w6, w4w9, w7h are included in the set of edges. Let P =xabzcdy, R1 = sefbaxw1w2 . . . w9ydchgt, andR2 = sefw3w2w1w6w5w4w9w8w7hgt.(See Fig. 7.5.) This graph has the following property.

x y

z

R1

s t

x y

P

z

s t

R2

Figure 7.5: Graphs L1 and L2. Paths P1, P2, R1, R2 and P are shown by thicklines

Lemma 7.2.11 Let G be a Hamiltonian graph such that G[V ′] is isomorphic toL2. Furthermore, if all edges in E(G) \ E(G[V ′]) incident to V ′ are incident tothe copies of the vertices x, y, z, s, t in V ′, then every Hamiltonian cycle in Gincludes either the path R1, or two paths P and R2 as segments.

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The lemma easily follows from the presence of degree 2 vertices in the graphL2, since for any such a vertex, it and adjacent vertices have to belong to onesegment of a Hamiltonian path.

Now we are ready to describe our reduction. Let (G, c) be a capacitated graphwith the vertex set v1, . . . , vn, m edges, and let k be a positive integer. Forevery vertex vi, four vertices ai, bi, ci and wi are introduced and the vertices biand ci are joined by c(vi)+1 paths of length two. Let Ci denote the set of middlevertices of these paths, and Xi = Ci∪ai, bi, ci. Then a copy Li

2 of the graph L2

with z = wi is added and vertices x and y of this gadget are joined by edges to ai

and bi respectively. By si and ti we denote the vertices s and t of Li2. For every

ordered pair vi, vj such that vivj ∈ E(G), a copy Lij2 of L2 is attached with

z = wj and vertices x and y made adjacent to all the vertices of Ci. The verticescorresponding to s and t are called sij and tij in Lij

2 . Furthermore, let xij andyij denote the vertices corresponding to x and y in Lij

2 . The path correspondingto P in Li

2 is called P i. Similarly, the path corresponding to P , R1 and R2 arecalled P ij, Rij

1 and Rij2 respectively in Lij

2 . Denote the obtained graph by G′(c).(See Fig. 7.6 for an illustration.)

In the next step we add two vertices g and h which are joined by∑n

i=1(c(vi)+4) + n+ 2m+ 1 paths of length two. Let Y be the set of middle vertices of thesepaths. All vertices si, ti, sij and tij are joined by edges with all vertices of Y .For every vertex r such that r ∈ Xi (recall Xi = Ci ∪ ai, bi, ci), i ∈ 1, . . . , n,a copy Lr

1 of L1 with z = r is attached and the vertices x, y of this gadget arejoined to all vertices of Y . We let xr and yr denote the vertices corresponding tox and y in Lr

1. Similarly P r1 and P r

2 denotes paths in Lr1 corresponding to P1 and

P2 respectively.

ai bi ci aj bj cj

wi wjw1 wn

a1 cn

z

x y

s t

L2

Figure 7.6: Graph G′(c)

Finally we add k+1 vertices, namely p1, . . . , pk+1, and make them adjacentto all the vertices ai, ci : 1 ≤ i ≤ n and to g and h. Let this resultant graph beH . The construction of H can easily be done in time polynomial in n and m.

Lemma 7.2.12 A graph (G, c) has a capacitated dominating set of size at mostk if and only if H has a Hamiltonian cycle.

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Proof. Let S be a capacitated dominating set of size at most k in (G, c) with thecorresponding dominating mapping f . Without loss of a generality we assumethat |S| = k and S = v1, . . . , vk. The Hamiltonian cycle we are trying toconstruct is naturally divided into k+ 1 parts by the vertices p1, . . . , pk+1. Weconstruct the Hamiltonian cycle starting from the vertex p1. Assume that thepart of the cycle up to the vertex pi is already constructed. We show how toconstruct the part from pi to pi+1. We include the edge piai in it. We add to thecycle the path P i and two edges, which join the endpoints of Pi with ai and bi.Let J = j : f(vj) = vi. If J = ∅ then a bi − ci-path of length two which goesthrough one vertex of Ci is included in the cycle. Otherwise all paths P ij for j ∈ Jare included in the cycle as follows. We consider the paths P ij in the increasingorder of indices in J and add them to the cycle. We take the first path say P ij′

and attach xij′ and yij′ to a pair of vertices j1, j2 in Ci. Suppose iteratively wehave included first l ≥ 1 paths in J , and the lth path is incident to some jl, jl+1in Ci, now we attach the (l + 1)th path by attaching xit of this to jl+1 and yit ofthis to jl+2, where jl+2 is a new vertex not incident to any previously includedpaths. We can always find such a vertex as |J | ≤ c(vi) = |Ci|−1. Now we includethe edge bij1 and j|J |+1ci. Finally we include the edge cipi+1.

When the vertex pk+1 is reached we move to the set Y . Note that at this stageall vertices w1, . . . , wn are already included in the cycle. We start by includingthe edge pk+1g. We will add following segments to the cycle an connect themappropriately.

• For every Li2 we add the path Ri

1 to the cycle if P i was not included to it,and include the path Ri

2 otherwise. The number of such paths is n.

• Similarly, for every Lij2 , the path Rij

1 is added to the cycle if P ij was notincluded to it, else the path Rij

2 is added. Note that 2m such paths areincluded to the cycle.

• For every vertex r such that r ∈ Xi for some i ∈ 1, . . . , n, the path P r2 is

included in the cycle if r is already included in the constructed part of thecycle, else the path P r

1 is added. Clearly, we add∑n

i=1(c(vi) + 4) paths.

Finally the total number of paths we will add is∑n

i=1(c(vi)+4)+n+2m = |Y |−1.We add the segments of the paths mentioned with the help of vertices in Y , in theway we added the paths P ij with the help of vertices in Ci. Let the end points ofthe resultant joined path be q1, q2. Notice that (a) q1, q2 ∈ Y and (b) this pathinclude all the vertices of Y . Now we add edges gq1, q2h and hp1. This completesthe construction of the Hamiltonian cycle.

For the reverse direction of the proof, we assume that we have been given C,a Hamiltonian cycle in H . Let S =

vi

∣∣ pjai ∈ E(C), aips /∈ E(C), j 6= s, forsome j ∈ 1, 2, . . . , k + 1

. We prove that S is a capacitated dominating set in

G of cardinality at most k. We first argue about the size of S, clearly its size is

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upper bounded by k + 1. To argue that it is at most k, it is enough to observethat by Lemmas 7.2.10 and 7.2.11 either pjg, or pjh must be in E(C) for somej ∈ 1, . . . , k + 1. Now we show that S is indeed a capacitated dominating set.Our proof is based on following observations.

• Every vertex wj, either appear in a vertex segment, that is P j, or an edgesegment, that is, P ij for some j ∈ 1, . . . , n in C.

• If some P ij appear as a segment in C, then from the gadgets Lbi

1 and Lci

1 thepaths P bi

2 and P ci

2 are part of C. Hence the only way to include bi in C isby using the edge incident to it from the gadget Li

2. This implies that fromthe gadget Li

2 we use the path P i and two edges, which join the endpointsof Pi with ai and bi.

• By Lemma 7.2.10 the cycle contains the edge which joins ai to some vertexin p1, . . . , pk+1.

Now given vj ∈ V (G) \ S, for the domination function f , we assign it to vi forwhich P ij is segment in C. Clearly vi ∈ S as by above observation there exits aj ∈ 1, 2, . . . , k + 1 such that pjai ∈ E(C), aips /∈ E(C) and j 6= s. For everyvi ∈ S, the set f−1(vi) contains at most c(vi) vertices as |Ci| = c(vi) + 1. Thisconcludes the proof.

The next lemma upper bounds the clique-width of the resulting graph H .

Lemma 7.2.13 If tw(G) ≤ t then cwd(H) ≤ 9·2max2t,24+12 and an expressiontree for H of width at most w = 9 · 2max2t,24 + 12 can be constructed in FPTtime.

Proof. We define c′(vi) = 0 for all i ∈ 1, 2, . . . , n and consider the graph G′(c′).It is easy to see that tw(G′(c′)) ≤ max2t + 1, |V (L2)| + 3 = max2t + 1, 25.By Theorem 1.2.1 cwd(G′(c′)) ≤ 3 · 2max2t,24, i.e. we can construct the labeledgraph G′(c′) by using at most l = 3 · 2max2t,24 labels α1, . . . , αl. Using l + 1additional labels β1, . . . , βl and γ1 we can ensure that all vertices si, ti, sij and tijare labeled by the label γ1, and only these vertices have label γ1 in the followingway. At the moment when such a vertex r labeled e.g. j is introduced, we labelit by the label βj , and then these labels are used in the operations in same way aslabels αj . Finally, all vertices labeled by these labels are relabeled γ1. Similarly,by using l + 1 more labels we assume that all vertices ai and ci are labeled bythe label γ2, and this label is used only for these vertices. Denote by di the onlyvertex in the set Ci in G′(c′). The graph G′(c) can be obtained from G′(c′) by thesubstitution of di by c(vi) + 1 vertices with same neighborhoods. This operationdoes not change clique-width, and cwd(G′(c)) ≤ 3l + 2.

Recall that for every vertex r ∈ Xi we add a copy of L1 with z = r. Weshow how to construct the obtained graph using no more than |V (L1)| + 1 = 8

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additional labels, in such a way that vertices xr and yr are labeled by the labelγ1. When a vertex r is introduced, we construct a copy of L1 using |V (L1)| extralabels making sure that the z in this copy gets r’s label. Then we relabel x andy by γ1, and the remaining |V (L1)| − 3 vertices are relabeled by an additionallabel, ζ , which acts as a “waste” label. We use 2 labels to construct the verticesg and h with |Y | paths of length two between them. Additionaly, we ensure thatat the end of this construction g and h are labeled with γ2, and that the verticesof Y are labeled by γ3.

Now, the join operation is done for vertices labeled γ1 and γ3. Now by usingone more label γ4, the vertices p1, p2, . . . , pk+1 are introduced, and the join oper-ation is performed on the labels γ2 and γ4. We used no more than 3l + 12 labelsto construct H , and cwd(H) ≤ 3l+ 12 ≤ 9 · 2max2t,24 +12. The second claim ofthe lemma is obvious.

Lemmas 7.2.12 and 7.2.13 together imply the following result.

Theorem 7.2.14 The Hamiltonian Cycle problem is W [1]-hard when param-eterized by clique-width. Moreover, this problem remains W [1]-hard even if theexpression tree is given.

7.2.4 Conclusion

Our results rule out FPT algorithms for Graph Coloring, Hamiltonian Cy-cle and Edge Dominating Set parameterized by cliquewidth unless FPT=W[1].We conclude the chapter with two open problems. The best known algorithmsfor these problems have running times on the form n2O(cwd(G))

, an interesting ques-tion is, can they be solved in nkO(1)

time? Another graph problem expressible inMSO2 but not MSO1 is the Max Cut problem. Is this problem also W[1]-hardparameterized by cliquewidth, or can an FPT algorithm be derived for it?

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Chapter 8

Kernelization Techniques

8.1 Meta-Theorems for Kernelization

One of the most important results in the area of kernelization is given by Alberet al. [5]. They provided the first kernel of linear size for the Dominating Setproblem on planar graphs. The work of Alber et al. [5] has triggered an explo-sion of papers on kernelization, and in particular on kernelization of problems onplanar graphs. Combining the ideas of Alber et al. [5] with problem specific datareduction rules, kernels of linear sizes were obtained for a variety of parameterizedproblems on planar graphs including Connected Vertex Cover, MinimumEdge Dominating Set, Maximum Triangle Packing, Efficient EdgeDominating Set, Induced Matching, Full-Degree Spanning Tree,Feedback Vertex Set, Cycle Packing, and Connected DominatingSet [5, 22, 23, 28, 79, 80, 87, 104, 112]

Most of the papers on linear kernels on planar graphs have the following ideain common: find an appropriate region decomposition of the input planar graphbased on the problem in question, and then perform problem specific rules toreduce the part of the graph inside each region. The first step towards the generalabstraction of all these algorithms was initiated by Guo and Niedermeier [79],who introduced the notion of “problems with distance property”. However, toprove that some problem admits a linear kernel on planar graphs, the approachof Guo and Niedermeier still requires the design of problem specific reductionrules. In this section we show that if a problem satisfies certain conditions,like expressibility in a certain kind of logic, then these reduction rules can beautomatically generated. We also extend the decomposition theorems provedby Guo and Niedermeier [79] to graphs of bounded genus, and show that ourreduction rules apply in bounded genus graphs as well. Our theorems unify andextend all previously known kernelization results for planar graph problems.

We say that a parameterized problem Π ⊆ Gg × N is compact if there existsan integer r such that for all (G = (V,E), k) ∈ Π there is an embedding of G

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86

into a surface Σ of Euler-genus at most g and a set S ⊆ V such that |S| ≤ r · k,Rr

G(S) = V , and k ≤ |V |r. Similarly, Π is quasi-compact if there exists an integerr such that for every (G, k) ∈ Π there is an embedding of G into a surface Σ ofEuler-genus at most g and a set S ⊆ V such that |S| ≤ r · k, tw(G \Rr

G(S)) ≤ rand k ≤ |V |r. Notice that if a problem is compact then it is also quasi-compact.

The first result proved in this section concerns a parameterized analogue ofgraph optimization problems where the objective is to find a maximum or mini-mum sized vertex or edge set satisfying a CMSO-expressible property. In partic-ular, the problems considered are defined as follows. In a p-min-CMSO graphproblem Π ⊆ Gg ×N, we are given a graph G = (V,E) and an integer k as input.The objective is to decide whether there is a vertex/edge set S of size at mostk such that the CMSO-expressible predicate PΠ(G, S) is satisfied. In a p-eq-CMSO problem the size of S is required to be exactly k and in a p-max-CMSOproblem the size of S is required to be at least k. The annotated version Πα ofa p-min/eq/max-CMSO problem Π is defined as follows. The input is a triple(G = (V,E), Y, k) where G is a graph, Y ⊆ V is a set of black vertices, and kis a non-negative integer. In the annotated version of a p-min/eq-CMSO graphproblem, S is additionally required to be a subset of Y . For the annotated versionof a p-max-CMSO graph problem S is not required to be a subset of Y , butinstead of |S| ≥ k we demand that |S ∩ Y | ≥ k.

Our results. For a parameterized problem Π ⊆ Gg ×N, let Π ⊆ Gg ×N denotethe set of all no-instances of Π. The first main result of this chapter is thefollowing theorem.

Theorem 8.1.1 Let Π ⊆ Gg×N be a p-min/eq/max-CMSO problem and eitherΠ or Π is compact. Then the annotated version Πα admits a cubic kernel if Πis a p-eq-CMSO problem and a quadratic kernel if Π is a p-min/max-CMSOproblem.

We remark that a polynomial kernel for an annotated graph problem Πα, is apolynomial time algorithm that given an input (G = (V,E), Y, k) of Πα, computesan equivalent instance (G′ = (V ′, E ′), Y ′, k′) of Πα such that |V ′| and k′ ≤ kO(1).Theorem 8.1.1 has the following corollary.

Corollary 8.1.2 Let Π ⊆ Gg × N be an NP-complete p-min/eq/max-CMSOproblem such that either Π or Π is compact and Πα is in NP . Then Π admits apolynomial kernel.

Theorem 8.1.1 and its corollary give polynomial kernels for all graph problemson surfaces for which such kernels are known. However, many problems in theliterature are known to admit linear kernels on planar graphs. Our next theoremunifies and generalizes all known linear kernels for graph problems on surfaces.To this end we utilize the notion of finite integer index.

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Theorem 8.1.3 Let Π ⊆ Gg × N has finite integer index and either Π or Π isquasi-compact. Then Π admits a linear kernel.

Theorems 8.1.1 and 8.1.3 will be proved in Section 8.1.3. The theorems aresimilar in spirit, yet they have a few differences. In particular, not every p-min/eq/max-CMSO graph problem has finite integer index. For an examplethe Independent Dominating Set problem is a p-min-CMSO problem, butit does not have finite integer index. Also, the class of problems that have finiteinteger index does not have a syntactic characterization and hence it takes somemore work to apply Theorem 8.1.3 than Theorem 8.1.1. On the other hand,Theorem 8.1.3 yields linear kernels, applies to quasi-compact problems and is notharder to apply than to design dynamic programming algorithms for graphs ofbounded treewidth.

To demonstrate the applicability of our results we show in Section 8.1.4 howour theorems lead to polynomial or linear kernels for a variety of problems. Forease of reference we provide a compendium of parameterized problems for whichpolynomial or linear kernels are derived in Section 8.1.5. We now proceed todevelop the tools necessary to prove Theorems 8.1.1 and 8.1.3.

8.1.1 Reduction Rules

In this section we give reduction rules for compact annotated p-min/eq/max-CMSO graph problems and quasi-compact parameterized problems having finiteinteger index. Our reduction rules have the following form:

If there is a constant size separator such that after its removal weobtain a connected component of unbounded size and of constanttreewidth, then we replace this component with something of constantsize.

The implementation of this rule depends on whether we are dealing with anannotated p-min-CMSO, p-eq-CMSO or p-max-CMSO problems, or whetherthe problem in question has finite integer index. Our reduction rules for annotatedp-min/eq/max-CMSO problems have three parts. In the first two parts we zeroin on an area to reduce, in the last part we perform the reduction. We now definethe notion of protrusions which formalizes the notion if a constant size separatorwhose removal creates a large connected component of constant treewidth.

Definition 8.1.4 [r-protrusion] Given a graph G = (V,E), we say that a setX ′ ⊆ V is an r-protrusion of G if |N(X ′)| ≤ r and tw(G[X ′ ∪N(X ′)]) ≤ r. Foran r-protrusion X ′, the vertex set X = X ′ ∪N(X ′) is an extended r-protrusion.The set X is the extended protrusion of X ′ and X ′ is the protrusion of X.

In the following discussion we only treat annotated p-min/eq/max-CMSO prob-lems where the set S being searched for is a set of vertices. The case where S isa set of edges can be dealt in an identical manner.

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Reduction for Annotated p-min-CMSO Problems

We now describe the reduction rule that we apply for annotated p-min-CMSOproblems. The technique employed in this section will act as a template forhow we handle the annotated p-eq/max-CMSO problems. Recall that in anannotated p-min-CMSO problem Πα we are given a graph G = (V,E) where asubset Y of the vertices of G is colored black and an integer k. The objective isto find a set S ⊆ Y of size at most k such that a fixed CMSO-definable propertyPΠ(G, S) holds. For our reduction rule, we are also given a sufficiently large r-protrusion X ′. In the first step of the reduction, we show that the set Y ∩X ′ canbe reduced to O(k) vertices without changing whether (G, k) is a yes-instance toΠα or not. In the second step we show that the r-protrusion X ′ can be coveredby O(k) r′-protrusions such that each r′-protrusion contains at most a constantnumber of vertices from Y . In the third and final step of the reduction rule,we replace the largest r′-protrusion with an equivalent, but smaller r′-boundariedgraph. We now provide the reduction rule for annotated p-min-CMSO problems.

Lemma 8.1.5 Let Πα be an annotated p-min-CMSO problem. Let G = (V,E)be a graph, Y ⊆ V be the set of black vertices and k be an integer. Let X bean extended r-protrusion of G. Then there is an integer c, and an O(|X|) timealgorithm, that computes a set of vertices Z ⊆ X ∩ Y with |Z| ≤ ck, such that ifthere exists a W ⊆ Y with PΠ(G,W ) and |W | ≤ k, then there exists a W ′ ⊆ Ywith PΠ(G,W ′), |W ′| ≤ k′, and W ′ ∩X ⊆ Z.

Proof. The algorithm starts by making a tree decomposition of G[X] of width atmost r, using the linear time algorithm to compute treewidth by Bodlaender [18].Now we add ∂(X) to each bag, and add one bag containing only the vertices in∂(X). The tree decomposition has width at most 2r as the bag size is at mostr + 1 + |∂(X)| ≤ 2r + 1.

Consider the following equivalence relation on subsets Q ⊆ X ∩ Y . We saythat Q ∼ Q′, if and only if for all R ⊆ V −X:

PΠ(G,Q ∪R) ⇔ PΠ(G,Q′ ∪R). (8.1)

The number of equivalence classes is bounded by a function of the treewidth [27,39] of G[X], and thus for fixed r, can be assumed to be bounded by a constant,say c. We would like to find a minimum sized representative of each of theequivalence classes. We describe an algorithm running in time O(|X|) to find thedesired set in each equivalence class. Let us consider an algorithm that solves anoptimization version of the problem

min|W | | W ⊆ Y ∧ PΠ(G,W )

on graphs of bounded treewidth, using a dynamic programming approach. For anexample, we can use the algorithm described by Borie et al. [27]. The algorithm of

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Borie et al. [27] computes for each equivalence class in Relation 8.1 (or, possibly,a refinement of the Relation 8.1) the minimum size of a set in the class. This isdone in a dynamic programming fashion, computing each value given a table ofthese values for the children of the bag in the tree decomposition. The runningtime is linear for fixed c. It is not hard to observe that we can also compute foreach equivalence class a minimum size set Q ⊆ X ∩ Y that belongs to the class.This can be done in linear time. If there are more than one minimum size sets ina class, then the algorithm just picks one.

Let Q denote that set of equivalence classes of Relation 8.1 and suppose thatfor each class q ∈ Q that is non-empty, we have a minimum size representativeQq. By the above argument we can find a minimum size representative Qq foreach class in O(|X|) time. Now, set

Z =⋃

q∈Q,|Qq|≤k

Qq .

Suppose now that there exists W ⊆ Y with PΠ(G,W ) and |W | ≤ k. Considerthe equivalence class q that contains X ∩W . Let Qq be the selected minimumsize representative of q. Consider the set W ′ = (W \ X) ∪ Qq. As PΠ(G, (W \X) ∪ (X ∩W )), we have that PΠ(G, (W \ X) ∪ Qc) = PΠ(G,W ′). Since, Qq isa minimum size representative from q, we have that |Qq| ≤ |W \ X|, and that|W ′| ≤ |W |. Finally, since the number of equivalence classes in Q is a function ofr and each representative Qq has size at most k, we have that |Z| = O(k). Thisproves the lemma.

Using Lemma 8.1.5, we change the set Y to (Y \X) ∪ Z. We now show howto exploit the fact that Z contains O(k) vertices.

Partitioning Protrusions: In the second step of the reduction rule we parti-tion the extended r-protrusion X into smaller r′-protrusions.

Lemma 8.1.6 Let G = (V,E) be a graph, Y ⊆ V be the set of black verticesof G and k be an integer. Furthermore, let X be an extended r-protrusion andZ = X ∩ Y . There is an O(|X|) time algorithm that finds O(|Z|) extended r′-protrusions X1, X2, . . . , Xℓ such that X = X1 ∪X2 ∪ . . .∪Xℓ and for every i ≤ ℓ,Z ∩Xi ⊆ ∂(Xi).

Proof. We start by making a nice tree decomposition of G[X], and adding ∂(X)to all bags. Now, we mark a number of bags. First, we mark the root bag andeach forget node where a vertex in Z is forgotten. As each vertex is forgotten atmost once in a nice tree decomposition, so far we have O(|Z|) marked bags. Now,mark each bag that is the lowest common ancestor of two marked bags, until wecannot marks in this fashion. Standard counting arguments for trees show that

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this operation at most doubles the number of marks. Hence, there are at mostO(|Z|) marked bags.

We now split X into X1, X2, . . . , Xℓ as follows: we take parts of the treedecomposition, with internally no marked bags, plus the marked bags at theborder. Note that each such part has at most two marked bags, each of size atmost 2r and thus the size of ∂(Xi) is at most 4r for every i ≤ ℓ. Note that bythe way we constructed the sets Xi, Z ∩Xi ⊆ ∂(Xi) for every i.

Reducing Protrusions: In the third phase of our reduction rule, we find aprotrusion to replace, and perform the replacement. Notice that this part ofthe reduction works both for annotated p-min-CMSO and for annotated p-eq-CMSO problems.

Lemma 8.1.7 Let Πα be an annotated p-min-CMSO or p-eq-CMSO problem.There is a fixed constant c depending only on Πα such that there is an algorithmthat given a graph G = (V,E) ∈ Gg, a set Y ⊆ V of black vertices, an integer kand an extended r-protrusion X with |X| > c such that Y ∩X ⊆ ∂(X), runs intime O(|X|), and produces a graph G∗ = (V ∗, E∗) ∈ Gg such that |V ∗| < |V | and(G∗, k) ∈ Πα if and only if (G, k) ∈ Πα.

Proof. For two t-boundaried graphs G1 and G2, we say that they are equivalentwith respect to a subset S of ∂(G1) = ∂(G2) if for every G3 = (V3, E3) and setS ′ ⊆ V3 we have that PΠ(G1⊕G3, S∪S ′) if and only if PΠ(G2⊕G3, S∪S ′). If G1

and G2 are equivalent with respect to S we say that G1 ∼S G2. The canonicalequivalence relation for CMSO properties with free set variables has finite in-dex [27, 39] and hence the number of equivalence classes of ∼S depends only on tfor every fixed https://www.easychair.org/login.cgi?a=a01749b032b2;iid=12978S ⊆ ∂(G1). We make a new equivalence relation defined on the set of t-boundariedgraphs belonging to a graph class G. Two t-boundaried graphs G1, G2 ∈ Gg areequivalent if

• for every t-boundaried graph G3, G1 ⊕G3 is legal if and only if G2 ⊕G3 islegal; and

• for every S ⊆ ∂(G1) = ∂(G2), G1 ∼S G2.

Now, ∼S has a finite number of equivalence classes for every S ⊆ ∂(G1) and theclass Gg is characterized by a finite set of forbidden minors. Hence the numberof equivalence classes in the equivalence relation defined above is a function of t.Let S be a set of r-boundaried graphs containing one smallest representative foreach equivalence class of the relation above. Let c the size of the largest graphin S. We have that X is a r-boundaried graph with boundary ∂(X). Let G1 bea graph in S such that G1 and G[X] are equivalent.

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Now we replace G[X] with G1 = (V1, E1) in the graph G, and let the resultinggraph be G∗ = (V ∗, E∗). Let Y1 be the set of black vertices in ∂(G1). We letY ∗ = Y \X ∪ Y1 be the set of black vertices in G∗. Since |X| > c, we have that|V1| < |X| and hence |V ∗| < |V |. It remains to prove that (G, k) ∈ Πα if andonly if (G∗, k) ∈ Πα. In one direction, suppose there is a set S ⊆ Y such thatPΠ(G, S) holds. Then S ∩ X ⊆ ∂(G[X]) and since G[X] and G1 are equivalentwith respect to the relation above we have that PΠ(G∗, S) holds. In the otherdirection, suppose PΠ(G∗, S) holds. Since Y ∗ ∩ V1 ⊆ ∂(G1) and G[X] and G1 areequivalent with respect to the relation above, we have that PΠ(G, S) holds. Thisconcludes the proof.

Lemmata 8.1.5, 8.1.6 and 8.1.7 together yield a reduction rule for all annotatedp-min-CMSO problems.

Lemma 8.1.8 Let Πα be an annotated p-min-CMSO problem. There is a fixedconstant c depending only on Πα such that there is an algorithm that given agraph G = (V,E) ∈ Gg, a set Y ⊆ V of black vertices, an integer k and anextended r-protrusion X with |X| > ck, runs in time O(|X|), and produces agraph G∗ = (V ∗, E∗) ∈ Gg such that |V ∗| < |V | and (G∗, k) ∈ Πα if and only if(G, k) ∈ Πα.

Proof. The algorithm starts by applying Lemma 8.1.5 to X, thus making allbut at most ak black vertices uncolored for some fixed constant a. By Lemma8.1.6, X = X1∪X2 . . .∪Xbk for some fixed constant b, where for every i, Xi is anextended 4r-protrusion such that Y ∩Xi ⊆ ∂(Xi). By the pigeon-hole principlesome Xi has size at least |X|/bk > c/b. Choose c such that c/b is sufficientlylarge to apply the algorithm in Lemma 8.1.7, and then apply Lemma 8.1.7 on theextended protrusion Xi. This concludes the proof.

Reduction for Annotated p-eq-CMSO Problems

In this section we give a reduction rule for annotated p-eq-CMSO problems.The rule is very similar to the one for the p-min-CMSO problems described inthe previous section. Therefore we only highlight the differences between the tworules in our arguments. The main difference between the two problem variantsis that we need to keep track of solutions of every fixed size between 0 and k,instead of just the smallest one in each class. Because of this we require theprotrusion to contain at least ck2 vertices instead of ck vertices, in order to beable to reduce it.

Lemma 8.1.9 Let Πα be an annotated p-eq-CMSO problem. There is a fixedconstant c depending only on Πα such that there is an algorithm that given agraph G = (V,E) ∈ Gg, a set Y ⊆ V of black vertices, an integer k and an

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extended r-protrusion X with |X| > ck2, runs in time O(k|X|), and produces agraph G∗ = (V ∗, E∗) ∈ Gg such that |V ∗| < |V | and (G∗, k) ∈ Πα if and only if(G, k) ∈ Πα.

Proof. We show that if Y ∩ X ≥ ak2 for some fixed constant k, then we canremove some vertices from Y preserving the answer. The proof proceeds almostas the proof of Lemma 8.1.5. The main difference is that now, instead of taking aminimum size representative from each equivalence class we consider all possiblesizes for S between 0 and k, for each equivalence class. That is, for each ℓ,0 ≤ ℓ ≤ k, and each equivalence class q, we make a set Qq,ℓ ⊆ X from the classwith |Qq,ℓ| = ℓ. Now, set

Z =⋃

q∈Q,0≤ℓ≤k,|Qq,ℓ|=ℓ

Qq,ℓ .

In the dynamic programming algorithm, we must also have one table entry foreach class and each size. This gives a running time of O(k|X|) for the first partof the reduction rule.

Next, we remove all vertices in X \ Z from Y and apply Lemma 8.1.6. Thisgives us X = X1 ∪X2 . . . ∪Xbk2 for some constant b, where for every i, Xi is anextended 4r-protrusion with Z ∩Xi ⊆ ∂(Xi).

By the pigeon-hole principle some Xi has size at least |X|/bk2 > c/b. Choosec such that c/b is sufficiently large to apply the algorithm in Lemma 8.1.7, andthen we apply Lemma 8.1.7 on the extended protrusion Xi. This concludes theproof.

Reduction for Annotated p-max-CMSO Problems

We now give a reduction rule for annotated p-max-CMSO problems. The ruleis still similar to the ones described in the two previous sections, but differs morefrom the p-min-CMSO problems than p-eq-CMSO did.

Lemma 8.1.10 Let Πα be an annotated p-max-CMSO problem. There is afixed constant c depending only on Πα such that there is an algorithm that givena graph G = (V,E) ∈ Gg, a set Y ⊆ V of black vertices, an integer k and anextended r-protrusion X with |X| > ck, runs in time O(|X|), and produces agraph G∗ = (V ∗, E∗) ∈ Gg such that |V ∗| < |V | and (G∗, k) ∈ Πα if and only if(G, k) ∈ Πα.

Proof. We again begin in a manner similar to the proof of Lemma 8.1.5. Themain ingredient in the proof of Lemma 8.1.5 is that for a given extended r-protrusion X, we consider the equivalence relation ∼ on subsets Q ⊆ X, wherewe demand that Q ∼ Q′ if and only if for all R ⊆ V −X: PΠ(Q∪R) ⇔ PΠ(Q′ ∪

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R). The number of equivalence classes of ∼ is bounded, but for maximizationproblems we need to keep the largest representative of each class. Hence we cannot guarantee that the union of all the representatives we store is bounded by afunction of k. To overcome this difficulty we use the expressive power providedby annotation. We compute a largest representative Qq ⊆ X for each equivalenceclass q of ∼. Then, we build a vertex set Z ⊆ X ∩ Y as follows. For eachnon-empty equivalence class q, if |Qq ∩ Y | ≤ k, then we add Qq ∩ Y to Z. If|Qq ∩ Y | > k, then we select arbitrarily a k-sized subset of Qq ∩ Y and add it toZ. Thus |Z| ≤ ak for some constant a. We remove all vertices in X \ Z from Y ,without changing the membership of (G, k) in Πα.

Next we apply Lemma 8.1.6 on X. This gives us X = X1 ∪ X2 . . . ∪ Xbk

for some constant b, where for every i, Xi is an extended 4r-protrusion withZ ∩Xi ⊆ ∂(Xi).

We now describe how to modify Lemma 8.1.7 so that it can also be appliedto annotated p-max-CMSO problems. For a set S ⊆ Y we define P ′

Π(G, S) asthere exists a set S ′ containing S such that PΠ(G, S ′) holds. Clearly (G, k) ∈ Πα

if and only if there is a set S ⊆ Y of size k such that P ′Π(G, S). If the relation

∼S in Lemma 8.1.7 is defined using P ′Π(G, S) instead of PΠ(G, S), the proof goes

through also for annotated p-max-CMSO problems.

By the pigeon-hole principle some Xi has size at least |X|/bk > c/b. Choosec such that c/b is sufficiently large to apply the algorithm in the modified versionof Lemma 8.1.7, and then apply the modified version of Lemma 8.1.7 to p-max-CMSO problems on the extended protrusion Xi. This concludes the proof.

Reductions Based on Finite Integer Index

In the previous sections we gave reduction rules for annotated p-min/eq/max-CMSO problems. These reduction rules, together with the results proved laterin this article will give quadratic or cubic kernels for the problems in question.However, for many problems we can in fact show that they admit a linear kernel.In this section we provide reduction rules for graph problems that have finiteinteger index. These reduction rules will yield linear kernels for the problemsthey apply to. We are now ready to prove the reduction lemma for problems thathave finite integer index.

Lemma 8.1.11 Let Π ⊆ Gg × N be problem with finite integer index in Gg suchthat either Π or Π is quasi-compact. There exists a constant c and an algorithmthat given a graph G = (V,E) ∈ G, an integer k and an extended r-protrusion Xin G with |X| > c, runs in time O(|X|) and returns a graph G∗ = (V ∗, E∗) ∈ Gg

and an integer k∗ such that |V ∗| < |V |, k∗ ≤ k and (G∗, k∗) ∈ Π if and only if(G, k) ∈ Π.

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Proof. Let S be a set of representatives for (Π, r) and let c = maxY ∈S |Y |.Similarly let S ′ be a set of representatives for (Π, 2r) and let c′ = maxY ∈S′ |Y |.If |X| > 3c′ we find an extended 2r-protrusion X ′ ⊆ X such that c′ < |X ′| ≤ 3c′

and work on X ′ instead of X. This can be done in time O(|X|) since G[X] hastreewidth at most r. From now on, we assume that |X| ≤ 3c′. This initial step isthe only step of the algorithm that does not work with constant size structures,and hence the running time of the algorithm is upper bounded by O(|X|). Thealgorithm proceeds as follows.

Because Π has finite integer index there is a graph H = (VH , EH) ∈ S suchthat H ≡Π G[X]. We show how to compute H from X. Let kmax = (6c′)p wherep is the smallest constant that satisfies the conditions for Π or Π being quasi-compact. For every G1 = (V1, E1) ∈ S, G2 = (V2, E2) ∈ S and k′ ≤ kmax wecompute whether (G1 ⊕G2, k

′) ∈ Π. For each such triple the computation can bedone in time O((|V1| + |V2|)p) since Π has finite integer index [25, 41]. Now, forevery G1 ∈ S and k′ ≤ (|X| + |V1|)p we compute whether (G[X] ⊕ G1, k

′) ∈ Π.When all these computations are done, the results are stored in a table.

It is not hard to see that H ≡Π G[X] if and only if there exists a constant csuch that for allG2 ∈ S and k′ ≤ kmax, (H⊕G2, k

′) ∈ Π ⇔ (G[X]⊕G2, k′+c) ∈ Π.

Also, c is the constant such that for all r-boundaried graphs G2 and and integersk′, (H ⊕ G2, k

′) ∈ Π ⇔ (G[X] ⊕ G2, k′ + c) ∈ Π. For each H ∈ C we can check

whether H ≡Π G[X] using this condition and the pre-computed table, and ifH ≡Π G[X], find the constant c.

After we have found a H ∈ S and the corresponding constant c, such thatH ≡Π G[X], we make G∗ from G by replacing the extended r-protrusion X withH . Also, we set k∗ = k − c. Since |X| > c and H has at most c vertices,|V ∗| < |V |. By the choice of H and c, (G∗, k∗) ∈ Π if and only if (G, k) ∈ Π.This concludes the proof.

8.1.2 Decomposition Theorems

Definition 8.1.12 We say that a graph G = (V,E) is (α, β, γ)-structured aroundS if |S| ≤ α and V can be partitioned into S,C1, C2, . . . , Cγ such that NG(Ci) ⊆ S,and max|NG(Ci)|, tw(G[Ci ∪NG(Ci)]) ≤ β, for every i = 1, . . . , γ.

Let G = (V,E) be a graph embedded in some surface Σ. A noose in G is aclosed curve N of Σ meeting only the vertices of G, we denote these vertices byVN = V ∩ N . We also define the length of N as the number of vertices it meetsand we denote it by |N |, that is, |N | = |VN |. The face-width of G is the minimumlength of a non-contractible noose in G. We denote the Euler genus of a surfaceΣ by eg(Σ).

Lemma 8.1.13 Let G = (V,E) be a graph in Gg and let S ⊆ V such thatBr

G(S) ≤ q. Then, tw(G) ≤ 4(2r + 1)√q + 2g + 8r + 12g.

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Proof. The result follows closely the argument of the proof of [44, Theorem 3.2]that, in turn, is based on [44, Lemma 3.1] about the distribution of every suchset S in the interior of a (ρ × ρ)-grid. As now we have graphs of higher genuswe have to apply [47, theorem 4.7] and find a lower bound to the size of S in thegraph obtained by a (ρ× ρ)-grid after adding O(g) edges.

From now on, we set f(r, g) = 4(2r + 1)√

2 + 2g + 8r + 12g.

Lemma 8.1.14 Let G = (V,E) be a graph, embedded in a surface Σ of Eulergenus g such that either g = 0 or the face-width of G is strictly greater than4r+2. Let S ⊆ V be a set containing at least 3 vertices where Br

G(S) = V . Thenthere exists a set S ′ ⊆ V such that S ⊆ S ′ and G is (r · (4r + 2) · (3|S| − 6 +6g), f(r, g), r · (3|S| − 6 + 6g))-structured around S ′.

Proof. We first need the following claim.Claim: Let G = (V,E) be a graph, embedded in a surface Σ of Euler genus gsuch that either g = 0 or the face-width of G is more than 4r + 2. Let S be aset of at least 3 vertices such that Br

G(S) = V . Then there is a collection R ofclosed subsets of Σ, also known as regions, such that

• If two sets in R have common points, then these points lay on their bound-aries.

• ⋃R∈RR ∩ V = V .

• The boundary of each R ∈ R is the union of two paths of length ≤ 2r + 1between two vertices of S called the anchors of R. We denote the set ofvertices on the boundary of R by bor(R).

• For each R ∈ R with u and v as anchors it holds that R∩ V ⊆ BrG(u, v).

• For each R ∈ R with u and v as anchors it holds that S ∩ R = u, v.• |R| ≤ r · (3|S| + 6g − 6).

The above claim follows from [79, Lemma 1] for the case where g = 0. For the sakeof completeness, we briefly present this proof together with its natural extensionfor embeddings of higher genus provided that the face-width of G is > 4r + 2.A collection R of closed subsets of Σ is constructed by a greedy algorithm asfollows: Start with an empty R and, as long as there are vertices not containedin some region R in R, find a area-maximal region R defined by two paths oflength ≤ 2r + 1 between two vertices in S (its anchors) that do not contain anyvertex in S and add it to R. Notice that such a region R always exists even fornon-planar embeddings, provided that the face-width ofG is bigger than 4r+2, asthis region is always inside a big enough disk of the surface. To prove the claimedupper bound on the size of R, consider a multigraph GR (again embedded in Σ)with vertex set S and there is an edge between u, v ∈ S whenever u and v are theanchors of some region R ∈ R. Notice that GR is also embedded in Σ and has

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face-width bigger than 1. This in turn implies that if two vertices u, v are joinedby many copies of the same edge in GR, then a pair of these edges will define adisk ∆ in Σ containing in its interior all other copies. Moreover, we may assumethat Σ \ ∆ contains a vertex w of S (this is necessary in case G has no planarembedding). We claim that the following thinness property holds: if there arex ≥ 2r copies of the edge u, v, then there exists a vertex w′ ∈ S laying in theinterior of one of the the x−1 ≥ 2r−1 area minimal disks ∆1, . . . ,∆x defined bythe copies of u, v inside ∆ (the order is chosen such that consecutive disks havecommon edges). We prove this using the argument of the proof of [79, Lemma1]. Assume to the contrary and consider the disk ∆r. Indeed, by area-maximallyof the choice of the regions in the above greedy procedure, it follows that ∆r

contains a vertex z ∈ V whose distance in G from u and v and every other vertexin S is bigger than r, a contradiction. This implies that there exists a vertexw′ ∈ S laying in the interior of one of the the x − 1 ≥ 2r − 1 area minimaldisks ∆1, . . . ,∆x defined by the copies of u, v inside ∆. Using the thinnessproperty of GR (see also [5, Lemma 5]) along with the Euler formula for graphsembedded in higher genus surfaces, we derive the claimed bound for |R| and theclaim follows.

To complete the proof of the lemma, we define S ′ =⋃

R∈R bor(R) ∩ V , thatis, S ′ contains all the vertices belonging to the boundary of the sets in R. Aseach such boundary is the union of two (u, v)-paths of length ≤ 2r + 1 whereu, v ∈ S, we have that such a boundary can have at most 4r + 2 vertices. As|R| ≤ r · (3|S| + 6g − 6), we obtain that |S ′| ≤ r · (4r + 2) · (3|S| − 6 + 6g). Foreach Ri ∈ R assign a set Ci containing each connected component C = (VC , EC)of G \S ′ for which NG(VC) ⊆ bor(R). If a connected component of G \S ′ can beassigned to more than one Ri, break ties arbitrarily so that Ci, i = 1, . . . , |R|forms a partition of the set of the connected components of G \ S ′. Notice thatfor each C ∈ Ci, NG(VC) ⊆ Ri ⊆ S ′ and if we set Ci = ∪C∈Ci

VC we have thatNG(Ci) ⊆ Ri ⊆ S ′ and thus |NG(Ci)| ≤ 4r + 2 ≤ f(r, g) (for i = 1, . . . , |R|).It remains to prove that if Ji = G[Ci ∪ NG(Ci)] then tw(Ji) ≤ f(r, g). Forthis, recall that R ∩ V ⊆ Br

G(u, v) and from Lemma 8.1.13, we obtain thattw(Ji) ≤ f(r, g), (for j = 1, . . . , |R|).

We are now in position to prove the following.

Lemma 8.1.15 Let G = (V,E) be a graph, embedded in a surface Σ of Eulergenus g and let S ⊆ V , |S| ≥ 3 such that Rr

G(S) = V for some r ≥ 0. Then thereexists a set S ′ such that S ⊆ S ′ and G is (h(r, g)·|S|, h(r, g), h(r, g)·|S|)-structuredaround S ′, where h(r, g) = O(rg).

Proof. We use induction on g. In particular, we prove that there exists a setS ′ ⊆ V such that G is (αg,|S|, βg, γg,|S|)-structured around S ′ where, for g = 0, weset α0,x = r · (4r+ 2) · (3x− 6), β0 = f(r, 0) and γ0,x = r · (3x− 6) and for g ≥ 1

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we have αg,x = (2g − 1) · r · 6(4r + 2)2 + r · (4r + 2) · (3x+ 6g − 6), βg = f(r, g)and γg,x = (2g−1) · r · 6(4r+2)+ r · (3x+6g−6). Once we have proved this, thelemma follows by suitably choosing the function h. For our proof we distinguishtwo cases:

Case 1. g = 0 or the embedding of G has face-width > 4r + 2. Then wedraw G together with its radial graph RG = (VR, ER) and denote it by G =G∪RG = (V ,E), the superposition of the two drawings. By the definition of theradial graph, it follows that Br

G(S) = V . Clearly, as G is a subgraph of G, the

result will follow if we prove that G is (αg,|S|, βg, γg,|S|)-structured around someset S ′ ⊆ VR. If G is embeddable in the sphere, then G is also embeddable in thesphere and if G is embeddable in a surface Σ of Euler genus g > 4r + 2 then theface-width of G is also greater than 4r + 2. Therefore, in either case we have allthe conditions required to apply Lemma 8.1.14 and hence the claim follows byapplying Lemma 8.1.14.

Case 2. There is a non-contractible noose N in Σ of length at most 4r+2. Thenwe split the graph along the vertices VN of the noose and we distinguish twosubcases depending if N is a surface separating or not.

Subcase 2.1. If N is a surface separating noose, then the splitting of the verticesof N creates two graphs G1 = (V1, E1) and G2 = (V2, E2) embedded in surfacesΣ1 and Σ2 respectively such that if eg(Σi) = gi, i = 1, 2 then g1 + g2 ≤ g andg1 · g2 > 0. Let Si consists of the vertices in S ∩ Vi and all the (duplicated)vertices met by N . Notice that Rr

Gi(Si) = Vi and that |S1|+ |S2| ≤ 2|N |+ |S| ≤

2(4r + 2) + |S|.By the induction hypothesis, Gi is (αgi,|Si|, βgi

, γgi,|Si|)-structured around someset S ′

i where Si ⊆ S ′i and gi > 1, i = 1, 2. Let G+ be the disjoint union of G1

and G2 and observe that G+ is (αg1,|S1| + αg2,|S2|,maxβg1, βg2, γg1,|S1| + γg2,|S2|)-structured around S ′

1 ∪ S ′2. Notice that

αg1,|S1| + αg2,|S2| ≤ (2g1 − 1 + 2g2 − 1) · r · 6(4r + 2)2 +

r · (4r + 2)(3|S1| + 3|S2| + 6g1 + 6g2 − 12)

≤ (2g1 + 2g2 − 2) · r · 6(4r + 2)2 +

r · (4r + 2)(3(2(4r + 2) + |S|) + 6g − 6)

≤ (2g1 + 2g2 − 1) · r · 6(4r + 2)2 − r · 6(4r + 2)2 +

r · 6(4r + 2)2 + r · (4r + 2) · (3|S| + 6g − 6)

≤ (2g − 1) · r · 6(4r + 2)2 + r · (4r + 2) · (3|S| + 6g − 6)

= αg,|S|

Similarly, we can show that γg1,|S1| + γg2,|S2| ≤ γg,|S| and, as maxβg1, βg2 ≤ βg,we conclude that G+ is (αg,|S|, βg, γg,|S|)-structured around S ′

1 ∪ S ′2. As all the

duplicated vertices of VN are in S1 ∪ S2, we can identify back these duplicated

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vertices occuring in S1 and S2 and obtain that G is (αg,|S|, βg, γg,|S|)-structuredaround some set S ′.

Subcase 2.2. If N is not a surface separating noose, then the splitting of thevertices of N creates a new graph G0 = (V0, E0) embedded in a surface Σ0 of Eulergenus g0 where g0 < g. Notice that if S0 is the set of vertices inG0 consisting of thenon-duplicated vertices of S and all the duplicated vertices, then Rr

G0(S0) = V0

and |S0| ≤ 2|N | + |S| ≤ 2(4r + 2) + |S|. From the induction hypothesis G0 is(αg0,|S0|, βg0, γg0,|S0|)-structured around some set S ′

0 where S0 ⊆ S ′0. Notice that

αg0,|S0| ≤ (2g0 − 1) · r · 6(4r + 2)2 + r · (4r + 2) · (3|S0| + 6g0 − 6)

≤ (2(g − 1) − 1) · r · 6(4r + 2)2 + r · (4r + 2) · (3(2(4r + 2) + |S|) + 6g − 6)

≤ (2g − 1) · r · 6(4r + 2)2 − 2 · r · 6(4r + 2)2 +

r · 6(4r + 2)2 + r · (4r + 2) · (3|S| + 6g − 6)

(2g − 1) · r · 6(4r + 2)2 + r · (4r + 2) · (3|S| + 6g − 6)

= αg,|S|

and similarly γg0,|S0| ≤ γg,|S|. As βg0 ≤ βg, we conclude thatG0 is (αg,|S|, βg, γg,|S|)-structured around S ′

0. As all the duplicated vertices of VN are in S0, we canidentify them back and deduce that G is (αg,|S|, βg, γg,|S|)-structured around someset S ′. This concludes the proof.

8.1.3 Kernels

Armed with the tools developed in the previous two sections, we are now readyto prove Theorems 8.1.1 and 8.1.3. We say that an instance (G′, k′) of a parame-terized problem Π is reduced with respect to a set Q of reduction rules if none ofthe reduction rules in Q can be applied to (G′, k′).

Proof of Theorem 8.1.1

Proof. We first give a proof for the case when Π is compact and Πα is anannotated p-min-CMSO problem. A proof for the case when Π is compact andΠα is an annotated p-eq/max-CMSO problem is identical.

We know that Π is compact and hence there exists an integer r such thatfor all (G = (V,E), k) ∈ Πα, there is an embedding of G into a surface of genusat most g, and a set S ⊆ V such that Rr

G(S) = V and |S| ≤ r · k. We showthat for all (G, k) ∈ Πα, the equivalent instance (G′ = (V ′, E ′), k), reduced withrespect to the reduction rule given by Lemma 8.1.8, has |V ′| = O(k2). Since(G′ = (V ′, E ′), k) ∈ Πα and Π is compact, there exists an embedding of G′ intoa surface of genus at most g and a set S ⊆ V ′ such that Rr

G′(S) = V ′. Henceby applying Lemma 8.1.15 we obtain a set S ′ such that G′ is (α, β, γ)-structured

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around S ′, where α, γ = O(rg|S|) and β = O(rg). This implies that V ′ can bepartitioned into S ′, C1, C2, . . . , Cγ such that NG′(Ci) ⊆ S ′, |NG′(Ci)| ≤ β andtw(G′[Ci ∪NG(Ci)]) ≤ β for every i ≤ γ. Observe that each Ci is a β-protrusionin G′ and |Ci ∪ NG(Ci)| ≤ ck, where c is a constant of Lemma 8.1.8, otherwisewe could have applied Lemma 8.1.8. This implies that

|V ′| ≤ |S ′| +γ∑

i=1

|Ci| = O

(rg|S|+

γ∑

i=1

ck

)= O(k2),

for some fixed g and r. Here constants hidden in big-Oh depend only on r and g.So given an input (G, k), if the size of the reduced graph is more than c∗k2 for

some constant c∗ then we return NO else we have G′ as the desired annotatedkernel for G.

Now we give a proof for the case when Π is compact and Πα is an annotatedp-max-CMSO problem. A proof for the case when Π is compact and Πα isan annotated p-min/eq-CMSO problem is similar. Towards this end, we claimthat for all (G, k) ∈ Π

α, the equivalent instance (G′ = (V ′, E ′), k), reduced with

respect to the reduction rule given by Lemma 8.1.10, has |V ′| = O(k2). Theproof for the claim is identical to the one we gave above to bound all the YESinstance for an annotated compact p-min-CMSO problem. So given an input(G, k), if the equivalent instance (G′ = (V ′, E ′), k), reduced with respect to thereduction rule given by Lemma 8.1.10, is more than c∗k2 for some constant c∗

then we return YES else we have G′ as the desired annotated kernel for G. Thereason we return YES is that if (G, k) would have a NO instance then the sizeof |V ′| ≤ c∗k2 as (G, k) ∈ Π

α.

Proof of Corollary 8.1.2

Proof. We know that Π is NP-complete and the annotated version Πα is in NP.So given an instance (G = (V,E), k), we apply Theorem 8.1.1 on the annotatedinstance (G = (V,E), V, k), that is we take V as Y , the set of black vertices. If weget YES or NO as an answer then we return the same. Else for (G′ = (V ′, E ′), k)of size polynomial in k, we apply polynomial time many to one reduction fromΠα to Π on G′ and obtain a graph G′′ = (V ′′, E ′′) ∈ Gg and an integer k′ suchthat |V ′′|, k′ ≤ kO(1) and (G′, k) ∈ Πα if and only if (G′′, k′) ∈ Π. This impliesthat in this case we have polynomial kernel for Π.

Proof of Theorem 8.1.3

The idea of the proof is that if a problem has finite integer index then an in-stance reduced with respect to the reduction rule given by Lemma 8.1.11 hasbounded radial distance. We first prove a lemma which will assist us in provingTheorem 8.1.3.

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Lemma 8.1.16 Let G = (V,E) be a graph embedded into a surface of genus atmost g and S ⊆ V such that tw(G \Rr

G(S)) ≤ r and all r-protrusions of G havesize at most m. Then there exists a S ′ ⊆ V such that |S ′| ≤ |S|+ g · (m+2r+ 1)and V = Rm+3r+1

G (S ′).

Proof. We prove the lemma using induction on Euler genus g of the graphG \ Rr

G(S). We distinguish two cases:

Case 1. g = 0 or the embedding of G \ RrG(S) has face-width > m + 2r + 1.

We claim that V = Rm+3r+1G (S), that is, S ′ = S. For this assume, towards a

contradiction, that x ∈ V \Rm+3r+1G (S) and consider the subgraph J of G induced

by the set Rm+2r+1G (x), that is, the vertices of G that are within radial distance

at most m+ 2r + 1 from x. Notice that J is a subgraph of G \RrG(S), therefore

tw(J) ≤ r. As either g = 0 or because the face-width of G is > m + 2r + 1, allthe vertices and edges of J are embedded inside a closed disk in Σ. Moreover,there exist m + 2r + 1 nested disjoint cycles C1, . . . , Cm+2r+1 with vertex setsV1, . . . , Vm+2r+1 respectively in RG such that, if ∆i is the closed disk with Ci as itsborder and contains x then i < j implies that ∆i ⊂ ∆j . For i = 1, . . . , m+2r+1,Vi contains vertices and faces whose radial distance from x, in G, is exactly i. Weneed the following claim.Claim. Each cycle of the radial graph RG that is entirely in (∆m+r+1 \ ∆m) andseparates S and x, has length > 2r.Proof. Indeed, if this is not the case for some cycle C with vertex set VC , thenL = V ∩ VC is a separator of G where |L| ≤ r and such that the connectedcomponent, say F , of G \ L that contains x is a subgraph of J . Then tw(F ) ≤tw(J) ≤ r and F is an r-protrusion of G. As F contains x and has more than mvertices, it is a contradiction to the assumption that all r-protrusions of G havesize at most m.

Applying the above claim to the cycles Cm+1, . . . , Cm+2r+1 we obtain thatthey all have length > 2r. We now construct an auxiliary graph R∗ by takingRG ∩ (∆m+r+1 \ ∆m), adding a vertex s adjacent to all vertices of Cm+1 andadding a vertex t adjacent to all vertices in Cm+2r+1. We claim that there is no(s, t)-separator in R∗ of size ≤ 2r. Indeed, such a separator would imply theexistence of a cycle C in RG of size ≤ 2r, a contradiction to the above claim. ByMenger’s theorem, it follows that there are > 2r disjoint (s, t)-paths in R∗ thatcorrespond to > 2r disjoint paths from the vertices of Cm+1 to the vertices ofCm+2r+1. The intersection of these paths with cycles Cm+1, . . . , Cm+2r+1, implythe existence of a (2r + 1) × (2r + 1)-grid as a minor of RJ , where RJ is theradial graph corresponding to J . Therefore, tw(RJ) > 2r (from [19, Lemma 88]),which, using [98, Lemma 3], implies that tw(J) > r, a contradiction.

Case 2. There is a non-contractible noose N , meeting the vertices of V in VN ,in G \ Rr

G(S) of length at most r′ = m + 2r + 1 (assume that G \ RrG(S) is

embedded in a surface Σ). Let S ′′ = S ∪ VN . Observe that tw(G \ RrG(S ′′)) ≤ r

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and all r-protrusions of G have size at most m. Furthermore let Σ′ be the surfacein which G \ Rr

G(S ′′) can be embedded. Then eg(Σ′) < eg(Σ) ≤ g. Hence byinduction hypothesis, there exists a set S ′ such that

|S ′| ≤ |S ′′|+(g−1)(m+2r+1) ≤ |S|+|VN |+(g−1)(m+2r+1) ≤ |S|+g(m+2r+1),

and V = Rm+3r+1G (S ′). This concludes the proof.

We are now in position to give the proof of Theorem 8.1.3.

Proof. Let us assume that Π is quasi-compact. Fix t = c∗rg where c∗ is aconstant to be defined later. Let (G′ = (V ′, E ′), k′), k′ ≤ k, be a reduced instancewith respect to reduction rule given by Lemma 8.1.11. Hence there is no extendedt-protrusion of size more than c, where c is a constant appearing in the statementof Lemma 8.1.11 and (G, k) ∈ Π if and only if (G′, k′) ∈ Π and G′ ∈ Gg. Hence,what remains to show is that |V ′| = O(k). The proof for this is similar to theone given for Theorem 8.1.1.

Now we show that if (G′ = (V ′, E ′), k′) ∈ Π then |V ′| ≤ O(k). Since Π isquasi-compact and (G′, k′) ∈ Π, there is an embedding of G into a surface of genusat most g and a set S ⊆ V ′ such that |S| ≤ r · k′ and tw(G′ \Rr

G′(S)) ≤ r. Sinceall t-protrusions of G′ are of size most c we have that all r-protrusions of G′ are ofsize at most c. Hence by Lemma 8.1.16 there exists a set S ′ such that |S ′| ≤ |S|+g ·(c+2r+1) and V ′ = R3r+c+1

G′ (S ′). Given S ′, we apply Lemma 8.1.15 and obtaina set S ′′ such that G′ is (α, β, γ)-structured around S ′′, where α, γ = O(rg|S ′|)and β = O(rg). This implies that V ′ can be partitioned into S ′′, C1, C2, . . . , Cγ

such that NG′(Ci) ⊆ S ′′, |NG′(Ci)| ≤ β and tw(G′[Ci ∪ NG′(Ci)]) ≤ β for everyi ≤ γ. Fix c∗ such that t ≥ β. This implies that G′[Ci∪NG′(Ci)] is a t-protrusionof G′ and hence its size is bounded by c. Now we are ready to bound the size ofV ′.

|V ′| ≤ |S ′′| +γ∑

i=1

|Ci| ≤ |S ′′| + γ · β · c = O(r3g3c2k) = O(k),

for fixed r, g and c. So given an input (G, k), if the size of the reduced graph ismore than c · k for some c, we return NO else we have G′ as the desired kernel.

The proof for the case when Π is quasi-compact is similar to the proof wegave for the case when Π was compact and Πα was an annotated p-max-CMSOproblem in Theorem 8.1.1. This concludes the proof.

8.1.4 Implications of Our Results

In this section we mention a few parameterized problems for which we can obtaineither polynomial or linear kernel using either Theorem 8.1.1 or Theorem 8.1.3.Various other problems for which we can obtain either polynomial or linear kernelsusing our results are mentioned in appendix.

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A Sufficient Condition for Finite Integer Index

We first give a sufficient condition which implies that a large class of p-min/max-CMSO problems has finite integer index. We prove it here for vertex versionsof p-min/max-CMSO problems that is if Π is a p-min/max-CMSO problemthen PΠ is a property of vertex sets. The edge version can be dealt in a similarmanner.

Let Π be a p-min-CMSO problem and Ft be the set of pairs (G = (V,E), S)where G is a t-boundaried graph and S ⊆ V . For a t-boundaried graph G =(V,E) we define the function ζG : Ft → N ∪ ∞ as follows. For a pair (G′ =(V ′, E ′), S ′) ∈ Ft, if there is no set S ⊆ V (S ⊆ E) such that PΠ(G⊕G′, S ∪ S ′)holds, then ζG((G′, S ′)) = ∞. Otherwise ζG((G′, S ′)) is the size of the smallestS ⊆ V (S ⊆ E) such that PΠ(G ⊕ G′, S ∪ S ′) holds. If Π is a p-max-CMSOproblem then we define ζG((G′, S ′)) to be the size of the largest S ⊆ V (S ⊆ E)such that PΠ(G⊕ G′, S ∪ S ′) holds. If there is no set S ⊆ V (S ⊆ E) such thatPΠ(G⊕G′, S ∪ S ′) holds then ζG((G′, S ′)) = ∞.

Definition 8.1.17 A p-min-CMSO problem Π is said to be strongly monotoneif there exists a function f : N → N such that the following condition is satisfied.For every t-boundaried graph G = (V,E), there is a subset S ⊆ V such that forevery (G′ = (V ′, E ′), S ′) ∈ Ft such that ζG((G′, S ′)) is finite, PΠ(G⊕G′, S ∪ S ′)holds and |S| ≤ ζG((G′, S ′)) + f(t).

Definition 8.1.18 A p-max-CMSO problem Π is said to be strongly monotoneif there exists a function f : N → N such that the following condition is satisfied.For every t-boundaried graph G = (V,E), there is a subset S ⊆ V such that forevery (G′ = (V ′, E ′), S ′) ∈ Ft such that ζG((G′, S ′)) is finite, PΠ(G⊕G′, S ∪ S ′)holds and |S| ≥ ζG((G′, S ′)) − f(t).

Lemma 8.1.19 Every strongly monotone p-min-CMSO and p-max-CMSO prob-lem has finite integer index.

Proof. We prove for p-min-CMSO problems, the proof for p-max-CMSO issimilar. Let Π be a monotone p-min-CMSO problem. Then PΠ is a finitestate property of t-boundaried graphs with a distinguished vertex set S [27, 39].In particular for every t, there exists a finite set S of pairs (G, S) such thatG = (V,E) is a t-boundaried graph and S ⊆ V such that the set S satisfiesthe following properties. For any t-boundaried graph G1 = (V1, E1) and setS1 ⊆ V1 there is a pair (G2 = (V2, E2), S2) ∈ S such that for every t-boundariedgraph G3 = (V3, E3) and set S3 ⊆ V3 we have that PΠ(G1 ⊕ G3, S1 ∪ S3) ⇐⇒PΠ(G2 ⊕G3, S2 ∪ S3). We fix such a set S.

For a t-boundaried graph G = (V,E) we define the signature ζSG : S →N ∪ ∞ of G to be ζG with domain restricted to S. We now argue that for

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any t-boundaried graph G, the maximum finite value of ζSG is at most f(t) largerthan the minimum finite value taken by ζSG. Since Π is strongly monotone thereexists a subset S of V that satisfies the conditions of Definition 8.1.17. Let (G′ =(V ′, E ′), S ′) ∈ S such that ζSG((G′, S ′)) is finite. Then PΠ(G⊕G′, S∪S ′) holds andhence ζSG((G′, S ′)) ≤ |S|. Furthermore the conditions of Definition 8.1.17 implythat |S|−f(t) ≤ ζSG((G′, S ′)) ≤ |S|. Hence the minimum and the maximum finitevalues of ζSG can differ by at most f(t). By the pigeon hole principle there is afinite set R of t-boundaried graphs such that for any t-boundaried graph G thereis a GR ∈ R and a constant cR depending only on the size of G and GR, suchthat for all (G′, S ′) ∈ S we have ζGR

((G′, S ′)) + cR = ζG((G′, S ′)).We now argue that R forms a set of representatives for (Π, t). Let G = (V,E)

be a t-boundaried graph and let GR = (VR, ER) ∈ R and cR be a constant suchthat for all (G′, S ′) ∈ S we have ζGR

((G′, S ′)) + cR = ζG((G′, S ′)). Let G′ =(V ′, E ′) be a t-boundaried graph and k be an integer such that (G⊕G′, k) ∈ Π.We argue that (GR⊕G′, k−cR) ∈ Π. Let Z ⊆ V ∪V ′ be a minimum size set suchthat PΠ(G⊕G′, Z) is satisfied, Z ′ = Z∩V ′ and ZG = Z\Z ′. Since (G⊕G′, k) ∈ Πwe have that |Z| ≤ k and hence ζG(G′, Z ∩ V ′) is finite. Let (GS , ZS) ∈ S bethe representative of (G′, Z ′). Then PΠ(G ⊕ GS , ZG ∪ ZS) holds and |ZG| =ζG(GS , ZS). Now we have that ζGR

((GS , ZS)) + cR = ζG((GS , ZS)). Hence, thereis a set SR ⊆ VR of size |ZG|−cR such that PΠ(GR⊕GS , SR∪ZS) holds. Then, since(GS , ZS) is the representative of (G′, Z ′) we have that PΠ(GR⊕G′, SR∪Z ′) holds.Now we have that |SR ∪Z ′| ≤ |SR|+ |Z ′| = |ZG| − cR + |Z ′| = |Z| − cR ≤ k− cR.This implies that (GR ⊕ G′, k − cR) ∈ Π. The proof for the other direction thatif (GR ⊕G′, k − cR) ∈ Π then (G⊕G′, k) ∈ Π is symmetric. This concludes thelemma.

Covering and Packing Problems

Minor Covering and Packing: We give below a few generic problems whichsubsumes many problems in itself and fit in our kernelization framework. Let Hbe a finite set of connected planar graphs.

Vertex-H-CoveringInput: A graph G = (V,E) ∈ Gg and a non-negative integer k.Question: Is there an S ⊆ V such that |S| ≤ k and G[V \ S] does not contain

any of the graphs in H as a minor?

Vertex-H-PackingInput: A graph G ∈ Gg and a non-negative integer k.Question: Does there exist k vertex disjoint subgraphs G1, . . . , Gk of G such

that each of them contains some graph in H as a minor?

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Lemma 8.1.20 Let H be a finite set of connected planar graphs and let Π1 denoteVertex-H-Packing. Then Vertex-H-Covering has finite integer index andis quasi-compact and Vertex-H-Packing has finite integer index and Π1 isquasi-compact.

Proof. Let Πv denote Vertex-H-Covering. We first show that Πv is quasicompact. If (G = (V,E), k) ∈ Πv then we know that there exists a set S ⊆ Vsuch that G[V \S] does not contain any graph in H as a minor. Now we show thatthe treewidth of G[V \ S] is at most a constant. To show this we need followingresults.

• Every planar graph H = (VH , EH) is a minor of the r × r grid, wherer = 14|VH | − 24 [123].

• For any fixed graph H , every H-minor free graph that does not contain aw × w grid as a minor has treewidth O(w) [45].

Let q = max|H| | H ∈ H and w = 14q − 24. Then observe that G[V \ S] doesnot contain w×w grid as a minor and hence tw(G[V \S]) ≤ O(w). This impliesthat Πv is quasi-compact.

Next we show that Π1 is quasi-compact. We first introduce some definitions.Given a graph G = (V,E), we define the covering number of G with respect tothe class H, covH(G), as the minimum k such that there exists S ⊆ V of sizek such that G[V \ S] does not contain any of the graphs in H as a minor. Thepacking number of G with respect to the class H, is defined as,

packH(G) = maxk | ∃ a partition V1, . . . , Vk of V such that

∀i∈1,...,kG[Vi] contains a graph in H as a minor.

Less formally, packH(G) ≥ k if G contains k vertex-disjoint minors in H. Weneed the following Erdos-Posa type of result shown in [67] for our purpose.

Claim 8.1.21 Let H be a finite set of connected planar graphs, q = max|H| | H ∈H and G be a non-trivial minor-closed graph class. Then there is a constant σG,q

depending only on G and q such that for every graph G ∈ G, it holds that

packH(G) ≤ coverH(G) ≤ σG,q · packH(G).

Using Claim 8.1.21 we show that Π1 is quasi-compact. Observe that if (G, k) ∈ Π1

and (G, k + 1) /∈ Π1, then by Claim 8.1.21, coverH(G) = O(k). Hence by anargument, similar to the one used for showing that Πv is quasi-compact, we havethat Π1 is quasi-compact.

It is known that Πv is a p-min-CMSO problem and Π1 is a p-max-CMSOproblem. Now using Lemma 8.1.19 we show that Πv is finite integer index. Weshow that Πv is strongly monotone. Given a t-boundaried graph G = (V,E),

105

with ∂(G) as its boundary, let S ′′ ⊆ V be a minimum set of vertices in G suchthat G[V \S ′′] does not contain any graph in H as a minor. Take S = S ′′∪∂(G).Now for any (G′ = (V ′, E ′), S ′) ∈ Ft such that ζG((G′, S ′)) is finite we have thatG ⊕ G′[(V ∪ V ′) \ (S ∪ S ′)] does not contain any graph in H as a minor and|S| ≤ ζG((G′, S ′)) + t. This proves that Πv is strongly monotone.

Next we show that Π1 is finite integer index. Given a t-boundaried graphG = (V,E), we define its signature, ζG, as follows. Let F qt

t ⊆ Gg be the set of allt-boundaried graph of size at most qt and ζG : F qt

t → N. For every G′ ∈ F qtt we

define ζG(G′) = packH(G⊕G′). For any G′, G′′ ∈ F qtt , |ζG(G′)−ζG(G′′)| ≤ t+qt,

as ζG(∅) ≤ ζG(G′) ≤ ζG(∅) + qt + t for all G′ ∈ F qtt . Hence, by the pigeon

hole principle there is a finite set R of t-boundaried graphs such that for anyt-boundaried graph G there is a GR ∈ R and a constant cR depending only onthe size of G and GR, such that ζGR

+ cR = ζG.We now argue that R forms a set of representatives for (Π1, t). Let G = (V,E)

be a t-boundaried graph and let GR = (VR, ER) ∈ R and cR be a constant suchthat ζGR

+cR = ζG. Let G′ = (V ′, E ′) be a t-boundaried graph and k be an integersuch that (G⊕G′, k) ∈ Π1. We argue that (GR⊕G′, k−cR) ∈ Π1. Let S be a setof k vertex disjoint minors in G ⊕ G′ and H be a minor in S which goes acrossor touch the boundary ∂(G). Now contract the part of this minor belonging tothe side of G′ as much close to the boundary ∂(G) as possible. Since the sizeof largest graph in H is at most q, we have that the part of the H belonging tothe side of G′ can be contracted to the border except for some q vertices, whichcould possibly be hanging out of the boundary. We do this for every minor in Swhich is going across. Let S ′ be the set of minors resulting after the contractionoperation has been performed. Now we take the boundary ∂(G) and all theminors of S ′ hanging out of it, and call this resulting graph G = (V , E). Observethat we can have at most t minors in S ′ which can hang out of the border asthese are vertex disjoint minors. Hence |V | ≤ qt and G is a t-boundaried graph.Now we know that ζGR

+ cR = ζG and hence ζGR(G) + cR = ζG(G). By definition

ζGR(G) = packH(G⊕G) = packH(G⊕G′), and hence (GR⊕G′, k−cR) ∈ Π1. The

proof for the other direction that if (GR⊕G′, k− cR) ∈ Π1, then (G⊕G′, k) ∈ Π1

is symmetric. This concludes that Π1 is finite integer index.

Vertex-H-Covering contains various problems as a special case, for ex-ample: (a) Feedback Vertex Set by taking H = where is a cycle oflength at most 3; deleting at most k vertices to obtain a graph of fixed treewidtht; deleting at most k vertices to obtain a graph into a graph class M, where M ischaracterized by a finite set of minors. Similarly Vertex-H-Packing containsproblem like Cycle Packing as a special case.

Subgraph Covering and Packing:

Let S be a finite set of connected graphs.

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Vertex-S-CoveringInput: A graph G ∈ Gg and a non-negative integer k.Question: Is there a S ⊆ V such that |S| ≤ k and G[V \ S] does not contain

any of the graphs in S as a subgraph?

We similarly define Edge-S-Covering by demanding S ⊆ E in the above defi-nition.

Vertex-S-PackingInput: A graph G ∈ Gg and a non-negative integer k.Question: Does there exists k vertex disjoint subgraphs G1, G2, ..., Gk in G

such that, for all i Gi is isomorphic to a graph in S.

We define Edge-S-Packing by demanding that G1, . . . , Gk be edge disjoint sub-graphs of G.

We can not show that Vertex/Edge-S-Covering or Vertex/Edge-S-Packing or there no instances are compact unless we do the following simplepreprocessing.Redundant Vertex and Edge Rule: Given an input (G = (V,E), k) to Vertex/Edge-S-Covering or Vertex/Edge-S-Packing remove all edges and vertices thatare not part of any subgraph isomorphic to any graph in S.

We can perform the Redundant Vertex and Edge Rule in O(|V | · |S|) time bylooking at a small ball around an edge e or a vertex v and check whether theball contains a subgraph isomorphic to a graph in S and contains the edge e orthe vertex v. This algorithm to check a subgraph isomorphic to a given graphcontaining a particular vertex or edge appears in a paper by Eppstein [55].

Lemma 8.1.22 Let S be a finite set of connected graphs and Π1 and Π2 cor-respond to Vertex-S-Packing and Edge-S-Packing respectively. Then thefollowing hold: (a) Vertex-S-Covering has finite integer index and is com-pact; (b) Vertex-S-Packing has finite integer index and Π1 is compact; (c)Edge-S-Covering is p-min-CMSO problem and is compact; and (d) Edge-S-Packing is p-max-CMSO problem and Π2 is compact.

Proof. Let Πv and Πe denote Vertex-S-Covering and Edge-S-Coveringrespectively. Let s = max|V ∗| | G∗ = (V ∗, E∗) ∈ S. Without loss of generalitywe assume that an input (G, k) to all these problems are reduced with respect toRedundant Vertex and Edge Rule.

We first show that these problems or their no instances are compact. Let(G, k) ∈ Πv then we know that there is a S ⊆ V such that |S| ≤ k and G[V \ S]does not contain any of the graphs in S as a subgraph and every vertex and edgeis in some subgraph in G which is isomorphic to a subgraph in S. This impliesthat every vertex in u ∈ V \S is in at most r = O(s) distance away from a vertexin S and hence Br

G(S) = V . We can similarly show that Πe is compact. Next we

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show that Π1 is compact. Observe that if (G, k) ∈ Π1 and (G, k + 1) /∈ Π1 thenwe know that we have a set Z of k vertex disjoint subgraphs in G where each ofthem is isomorphic to a subgraph in S. Take S as the union of all the verticesappearing in any of the subgraph in Z. Note that |S| ≤ s ·k. Observe that S hitsall the subgraphs isomorphic to a subgraph in S and hence Br

G(S) = V , wherer = O(s). This implies that Π1 is compact. We can similarly show that Π2 iscompact.

It is well known that Πv and Πe are p-min-CMSO problems while Π1 and Π2

are p-max-CMSO problems. The proof that Πv and Π1 are finite integer index issimilar to the proof given for Vertex-H-Covering and Vertex-H-Packing,where H is a finite set of connected planar graphs, have finite integer index inLemma 8.1.20. The only thing we need to replace is minors by subgraphs in thatproof.

Domination and its Variants

In the r-Dominating Set problem, we are given a graph G = (V,E), anda positive integer k, and the objective is to find a subset S ⊆ V such thatBr

G(S) = V and |S| ≤ k. For r = 1, if we demand that G[S] is connectedthen we get Connected Dominating Set. A problem is called q-ThresholdDominating Set if we demand that B1

G(S) = V and for all v ∈ (V \ S),|N(v) ∩ S| ≥ q. An independent set C of vertices in a graph G = (V,E) is anefficient dominating set (or perfect code) when each vertex not in C is adjacentto exactly one vertex in C. In Efficient Dominating Set problem we aregiven a graph G = (V,E) and a positive integer k and the objective is to find anefficient dominating set of size at most k.

Lemma 8.1.23 r-Dominating Set, Connected Dominating Set, Effi-cient Dominating Set and q-Threshold Dominating Set are compactand have finite integer index.

Proof. All these problems are compact by definition. To show that these prob-lems have finite integer index, we make use of Lemma 8.1.19. Clearly, they are p-min-CMSO problems. We now show that these problems are strongly monotone.We first show it for r-Dominating Set. Given a t-boundaried graphG = (V,E),with ∂(G) as its boundary, let S ′′ ⊆ V be a minimum r-dominating set of G. TakeS = S ′′∪∂(G). Now for any (G′ = (V ′, E ′), S ′) ∈ Ft such that ζG((G′, S ′)) is finitewe have that S ∪ S ′ is a r-dominating set and |S| ≤ ζG((G′, S ′)) + t. This provesthat r-Dominating Set is strongly monotone. Similarly, we can show that q-Threshold Dominating Set is strongly monotone by taking S = S ′′ ∪ ∂(G)where S ′′ ⊆ V is a minimum q-threshold dominating set of G. To show that Con-nected Dominating Set is strongly monotone we take S = S ′′ ∪ ∂(G) where

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S ′′ ⊆ V is a union of minimum connected dominating set for each connectedcomponents of G.

We now prove that Efficient Dominating Set has finite integer index. LetΠ be the Dominating Set problem and Π′ be the Efficient Dominating Setproblem. The property P (G) that G has an efficient dominating set (of any size)is expressible in CMSO and hence this property is finite state in t-boundariedgraphs. As argued in the above paragraph, Dominating Set has finite integerindex. Furthermore by a theorem of Bange et al. [14], if a graph G has an efficientdominating set, then the size of the minimum efficient dominating set is equal tothe size of the minimum dominating set of the graph. Hence for two t-boundariedgraphs G1, G2 if G1 and G2 are in the same equivalence class of P and G1 ≡Π G2

then G1 ≡Π′ G2. Hence Π′ has a finite set of representatives.

Problems on Directed Graphs

Our results also apply to problems on directed graphs of bounded genus. In thisdirection we mention three problems considered in the literature. In DirectedDomination [4] we are given a directed graph D = (V,A) and a positive integerk and the objective is to find a subset S ⊆ V of size at most k such that for veryvertex u ∈ V \ S there is a vertex v ∈ S such that (u, v) ∈ A. IndependentDirected Domination1 [83] takes as input a directed graph D = (V,A) anda positive integer k and the objective is to find a subset S ⊆ V of size at mostk such that S is an independent set and for every vertex u ∈ V \ S there is avertex v ∈ S such that (u, v) ∈ A. In the Minimum Leaf Out-branching [82]problem we are given a directed graph D = (V,A) and a positive integer k andthe objective is to find a rooted directed spanning tree (with all arcs directedoutwards from the vertices) with at least k internal vertices.

Lemma 8.1.24 Independent Directed Domination is a p-min-CMSO com-pact problem, Directed Domination is compact and has finite integer indexand Π=Minimum Leaf Out-branching is a p-max-CMSO problem and Πis compact.

Proof. Independent Directed Domination and Directed Dominationcan easily be seen to be p-min-CMSO problems and by their definition they arecompact. Directed Domination can be shown to have finite integer indexas follows: given a t-boundaried graph G = (V,E), with ∂(G) as its boundary,let S ′′ ⊆ V be a minimum directed dominating set of G. Take S = S ′′ ∪ ∂(G).Now for any (G′ = (V ′, E ′), S ′) ∈ Ft such that ζG((G′, S ′)) is finite we have that

1In literature it is known as “Kernels”. We call it differently here to avoid confusion withproblem kernels.

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S ∪ S ′ is a directed dominating set and |S| ≤ ζG((G′, S ′)) + t. This proves thatDirected Dominating Set is strongly monotone.

Let Π be the Minimum Leaf Out-branching problem. Observe that Π isa p-max-CMSO problem. The set of no instances can be seen to be compactby observing the fact that if (G, k) ∈ Π and (G, k + 1) /∈ Π then G has an out-branching with exactly k internal vertices with all other vertices being its leaves.This implies that Π is compact.

Finally by applying Theorem 8.1.3 together with Lemmata 8.1.19, 8.1.20,8.1.22, 8.1.23, and 8.1.24 we get the following corollary.

Corollary 8.1.25 For g ≥ 0, Feedback Vertex Set, Edge DominatingSet, Vertex Cover, Dominating Set, r-Dominating Set, q-ThresholdDominating Set, Connected Dominating Set, Directed Domination,Connected Vertex Cover, Minimum-Vertex Feedback Edge Set, In-duced Matching, Minimum Maximal Matching, Efficient DominatingSet, Independent Set, Induced d-Degree Subgraph, Min Leaf Span-ning Tree, Triangle Packing, Cycle Packing, Cycle Domination,Maximum Full-Degree Spanning Tree, Vertex-H-Packing, Vertex-H-Covering, Vertex-S-Covering and Vertex-S-Packing admit a linearkernel on graph of genus at most g.

Corollary 8.1.25 unifies and generalizes results presented in [4, 5, 22, 23, 28,69, 79, 80, 87, 104, 112]. By applying Theorem 8.1.1, Corollary 8.1.2, andLemmata 8.1.20, 8.1.22, 8.1.23, and 8.1.24, we get the following corollary forproblems which are not finite integer index.

Corollary 8.1.26 For g ≥ 0, Independent Dominating Set, Indepen-dent Directed Domination, Minimum Leaf Out-branching, Edge-S-Covering and Edge-S-Packing admit a polynomial kernels on graphs of genusat most g.

8.1.5 Practical considerations:

Our meta-theorems provide a simple criteria to decide whether a problem admitsa polynomial or linear kernel. Our proofs are constructive and essentially providemeta-algorithms that construct kernels for problems in an automated way. Ofcourse, it is expected that for concrete problems, tailor-made kernels will givebetter bounds. Indeed, many of the published proofs for concrete problems havemuch smaller constant factors than would follow from a direct application ofour proofs. However, our approach might be useful for computer aided designof kernelization algorithms: with a Myhill-Nerode approach, we can get a setof rules that transform each region to the size of some minimum representative,and we can let the computer calculate such sizes or even output the reductions

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of the corresponding kernel. This seems an interesting (and far from trivial)algorithm-engineering problem.

In general, finding linear kernels with small constant factors for concrete prob-lems on planar graphs or graphs with small genus remains a worthy topic offurther research.

Compendium of Problems Considered in Section 8.1

Problems that have Finite Integer Index and Quasi-Compactness Prop-erty – Linear Kernels Dominating Set, r-Dominating Set, q-ThresholdDominating Set, Cycle Domination, Vertex Cover, Feedback Ver-tex Set, Connected Dominating Set, Connected r-Dominating Set,Connected Vertex Cover, Minimum-Vertex Feedback Edge Set, EdgeDominating Set, Minimum Maximal Matching, Efficient DominatingSet, Vertex-H-Covering, Vertex-S-Covering, Almost-Outerplanar,Clique-Transversal.

Problems that have Finite Integer Index and No Instances havingQuasi-Compactness Property – Linear Kernels Independent Set, In-duced d-Degree Subgraph, Min Leaf Spanning Tree, Induced Match-ing, Triangle Packing, Cycle Packing, Maximum Full-Degree Span-ning Tree, Vertex-H-Packing, Vertex-S-Packing.

Problems that are not covered above and are p-min/eq/max-CMSOproblems having Compactness Property – Polynomial Kernels Inde-pendent Dominating Set, Independent Directed Domination, Mini-mum Leaf Out-branching, Edge-S-Covering, Edge-S-Packing.

Problems that have Finite Integer Index but neither the problem norits no-instances satisfy Quasi-Compactness Property Minimum Par-tition Into Cliques, Hamiltonin Path Completion.

Problems that do not have Finite Integer Index Longest Path, LongestCycle, Maximum Cut, Minimum Covering by Cliques, IndependentDominating Set, Minimum Leaf Out-branching.

8.2 Turing Kernels

As we have seen in Chapter 5, it is not always possible to obtain polynomialkernels for all FPT problems. However, the traditional notion of kernelizationdoes not seem to capture all feasible variants of polynomial time preprocessing.

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For instance, the results from Chapter 5 can not be used to rule out a “cheatkernelization algorithm” that reduces the input instance (I, k) to |I| indepen-dent kernels of size O(k). That is, such an algorithm would output instances(X1, k

′) . . . (X|I|, k′) such that (I, k) is a yes instance if and only if (Xi, k

′) is ayes instance for some i, and if |Xi| = O(k) for every i. While an algorithm likethis is not as useful as an actual kernelization algorithm, it could still turn outto be quite helpful in practice. In particular problems admitting such a “cheatkernel” would be very amenable for parallel algorithms. In 2008, Guo and Fel-lows [16] asked whether there exist problems that admit such “cheat” polynomialkernels, but provably do not admit polynomial a kernel unless PH=Σ3

p. The re-sults in this section, together with the results in Section 9.1, give an affirmativeanswer to this question.

We now formally define the notion of Turing kernelization. Our notion issomewhat more general than the “cheat kernel” notion proposed by Guo andFellows in [16]. The reason we find this notion more appropriate is that eventhough our notion is more general, problems that admit turing kernels have almostas good algorithmic properties as the problems that admit “cheat kernels”. Inorder to define turing kernels we first define the notion of a t-oracle.

Definition 8.2.1 A t-oracle for a parameterized problem Π is an oracle thattakes as input (I, k) with |I| ≤ t, k ≤ t and decides whether (I, k) ∈ Π inconstant time.

Definition 8.2.2 A parameterized problem Π is said to have g(k)-sized turingkernel if there is an algorithm which given an input (I, k) together with a g(k)-oracle for Π decides whether (I, k) ∈ Π in time polynomial in |I| and k.

Observe that the traditional notion of kernelization is a special case of turingkernelization. In particular, a kernel is equivalent to a turing kernel where thekernelization algorithm is only allowed to make one oracle call and must returnthe same answer as the oracle.

8.2.1 The Rooted k-Leaf Out-Branching Problem

The k-Leaf Out-Branching problem is to find an out-branching, that is arooted oriented spanning tree, with at least k leaves in a given digraph. Theproblem has recently received much attention from the viewpoint of parame-terized algorithms [9, 8, 26, 94, 40]. Here, we take a kernelization based ap-proach to the k-Leaf-Out-Branching problem. We give the first polyno-mial kernel for Rooted k-Leaf-Out-Branching, a variant of k-Leaf-Out-Branching where the root of the tree searched for is also a part of the input.Our kernel has cubic size and is obtained using extremal combinatorics. Ourkernel for Rooted k-Leaf-Out-Branching yields a “cheat kernel” for the

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k-Leaf-Out-Branching problem. In Section 9.1 we show that the k-Leaf-Out-Branching problem does not admit a polynomial kernel unless PH=Σ3

p.These two results put together dempnstrate that turing kernelization indeed ismore powerful than kernelization in the traditional sense. We now give all thedata reduction rules we apply on the given instance of Rooted k-Leaf Out-Branching to shrink its size.

Proposition 8.2.3 ([94]) Let D be a digraph and r be a vertex from which everyvertex in V (D) is reachable. Then if we have an out-tree rooted at r with k leavesthen we also have an out-branching rooted at r with k leaves.

RedRule 8.2.1.1 [Reachability Rule] If there exists a vertex u which is discon-nected from the root r, then return No.

For the Rooted k-Leaf Out-Tree problem, Rule 8.2.1.1 translates intofollowing: If a vertex u is disconnected from the root r, then remove u and allin-arcs to u and out-arcs from u.

RedRule 8.2.1.2 [Useless Arc Rule] If vertex u disconnects a vertex v from theroot r, then remove the arc vu.

Lemma 8.2.4 Reduction Rules 8.2.1.1 and 8.2.1.2 are correct.

Proof. If there exists a vertex which can not be reached from the root r thena digraph can not have any r-out-branching. For Reduction Rule 8.2.1.2, allpaths from r to v contain the vertex u and thus the arc vu is a back arc in anyr-out-branching of D.

RedRule 8.2.1.3 [Bridge Rule] If an arc uv disconnects at least two verticesfrom the root r, contract arc uv.

Lemma 8.2.5 Reduction Rule 8.2.1.3 is correct.

Proof. Let the arc uv disconnect at least two vertices v and w from r and let D′

be the digraph obtained from D by contracting the arc uv. Let T be an r-out-branching of D with at least k leaves. Since every path from r to w contains thearc uv, T contains uv as well and neither u nor v are leaves of T . Let T ′ be thetree obtained from T by contracting uv. T ′ is an r-out-branching of D′ with atleast k leaves.

In the opposite direction, let T ′ be an r-out-branching of D′ with at least kleaves. Let u′ be the vertex in D′ obtained by contracting the arc uv, and let x bethe parent of u′ in T ′. Notice that the arc xu′ in T ′ was initially the arc xu beforethe contraction of uv, since there is no path from r to v avoiding u in D. We make

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p1 u v p8pin

p1 p4 p5 p8

pout

p2 p3 p6 p7

Figure 8.1: An Illustration of Reduction Rule 8.2.1.5.

an r-out-branching T of D from T ′, by replacing the vertex u′ by the vertices uand v and adding the arcs xu, uv and arc sets vy : u′y ∈ A(T ′) ∧ vy ∈ A(D)and uy : u′y ∈ A(T ′) ∧ vy /∈ A(D). All these arcs belong to A(D) because allout-neighbors of u′ in D′ are out-neighbors either of u or of v in D. Finally, u′

must be an inner vertex of T ′ since u′ disconnects w from r. Hence T has at leastas many leaves as T ′.

RedRule 8.2.1.4 [Avoidable Arc Rule] If a vertex set S, |S| ≤ 2, disconnects avertex v from the root r, vw ∈ A(D) and xw ∈ A(D) for all x ∈ S, then deletethe arc vw.

Lemma 8.2.6 Reduction Rule 8.2.1.4 is correct.

Proof. Let D′ be the graph obtained by removing the arc vw from D and letT be an r-out-branching of D. If vw /∈ A(T ), T is an r-out-branching of D′,so suppose vw ∈ A(T ). Any r-out-branching of D contains the vertex v, andsince all paths from r to v contain some vertex x ∈ S, some vertex u ∈ S isan ancestor of v in T . Let T ′ = (T ∪ uw) \ vw. T ′ is an out-branching of D′.Furthermore, since u is an ancestor of v in T , T ′ has at least as many leaves asT . For the opposite direction observe that any r-out-branching of D′ is also anr-out-branching of D.

RedRule 8.2.1.5 [Two Directional Path Rule] If there is a path P = p1p2 . . . pl−1pl

with l = 7 or l = 8 such that

• p1 and pin ∈ pl−1, pl are the only vertices with in-arcs from the outside ofP .

• pl and pout ∈ p1, p2 are the only vertices with out-arcs to the outside ofP .

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• The path P is the unique out-branching of D[V (P )] rooted at p1.

• There is a path Q that is the unique out-branching of D[V (P )] rooted at pin.

• The vertex after pout on P is not the same as the vertex after pl on Q.

Then delete R = P \ p1, pin, pout, pl and all arcs incident to these vertices fromD. Add two vertices u and v and the arc set poutu, uv, vpin, plv, vu, up1 to D.

Notice that every vertex on P has in-degree at most 2 and out-degree at most2. Figure 8.1 gives an example of an application of Reduction Rule 8.2.1.5.

Lemma 8.2.7 Reduction Rule 8.2.1.5 is correct.

Proof. Let D′ be the graph obtained by performing Reduction Rule 8.2.1.5 to apath P in D. Let Pu be the path p1poutuvpinpl and Qv be the path pinplvup1pout.Notice that Pu is the unique out-branching of D′[V (Pu)] rooted at p1 and thatQv is the unique out-branching of D′[V (Pu)] rooted at pin.

Let T be an r-out-branching of D with at least k leaves. Notice that since P isthe unique out-branching of D[V (P )] rooted at p1, Q is the unique out-branchingof D[V (P )] rooted at pin and p1 and pin are the only vertices with in-arcs fromthe outside of P , T [V (P )] is either a path or the union of two vertex disjointpaths. Thus, T has at most two leaves in V (P ) and at least one of the followingthree cases must apply.

1. T [V (P )] is the path P from p1 to pl.

2. T [V (P )] is the path Q from pin to pout.

3. T [V (P )] is the vertex disjoint union of a path P that is a subpath of Prooted at p1, and a path Q that is a subpath of Q rooted at pin.

In the first case we can replace the path P in T by the path Pu to get anr-out-branching of D′ with at least k leaves. Similarly, in the second case, wecan replace the path Q in T by the path Qv to get an r-out-branching of D′

with at least k leaves. For the third case, observe that P must contain pout sincepout = p1 or p1 appears before pout on Q and thus, pout can only be reached fromp1. Similarly, Q must contain pl. Thus, T \R is an r-out-branching of D \R. Webuild an r-out-branching T ′ of D′ by taking T \ R and letting u be the child ofpout and v be the child of pl. In this case T and T ′ have same number of leavesoutside of V (P ) and T has at most two leaves in V (P ) while both u and v areleaves in T ′. Hence T ′ has at least k leaves.

To show the other direction, let T ′ be an r-out-branching of D′ with at leastk leaves. Notice that since Pu is the unique out-branching of D′[V (Pu)] rootedat p1, Qv is the unique out-branching of D′[V (Pu)] rooted at pin and p1 and pin

are the only vertices with in-arcs from the outside of V (Pu), T′[V (Pu)] is either a

path or the union of two vertex disjoint paths. Thus, T ′ has at most two leavesin V (Pu) and at least one of the following three cases must apply.

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1. T ′[V (Pu)] is the path Pu from p1 to pl.

2. T ′[V (Pu)] is the path Qv from pin to pout.

3. T ′[V (Pu)] is the vertex disjoint union of a path Pu that is a subpath of Pu

rooted at p1, and a path Qv that is a subpath of Qv rooted at pin.

In the first case we can replace the path Pu in T ′ by the path P to get anr-out-branching of D with at least k leaves. Similarly, in the second case, wecan replace the path Qv in T ′ by the path Q to get an r-out-branching of D′

with at least k leaves. For the third case, observe that Pu must contain pout sincepout = p1 or p1 appears before pout on Qv and thus, pout can only be reachedfrom p1. Similarly, Qv must contain pl. Thus, T ′ \ u, v is an r-out-branchingof D′ \ u, v. Let x be the vertex after pout on P , and let y be the vertex afterpl on Q. Vertices x and y must be distinct vertices in R and thus there mustbe two vertex disjoint paths Px and Qy rooted at x and y, respectively, so thatV (Px)∪ V (Qy) = R. We build an r-out-branching T from (T ′ \ u, v)∪Px ∪Qy

by letting x be the child of pout and y be the child of pin. In this case T ′ and Thave the same number of leaves outside of V (P ) and T ′ has at most two leavesin V (Pu) while both the leaf of Pu and the leaf of Qv are leaves in T . Hence Thas at least k leaves.

We say that a digraph D is a reduced instance of Rooted k-Leaf Out-Branching if none of the reduction rules (Rules 1–5) can be applied to D. Itis easy to observe from the description of the reduction rules that we can applythem in polynomial time, resulting in the following lemma.

Lemma 8.2.8 For a digraph D on n vertices, we can obtain a reduced instanceD′ in polynomial time.

8.2.2 Bounding Entry and Exit Points

We show that any reduced no-instance of Rooted k-Leaf Out-Branchingmust have at most O(k3) vertices. In order to do so we start with T , a breadth-first search-tree (or BFS-tree for short) rooted at r, of a reduced instance D andlook at a path P of T such that every vertex on P has out-degree one in T .We now bound the number of endpoints of arcs with one endpoint in P and oneendpoint outside of P .

Let D be a reduced no-instance, and T be a BFS-tree rooted at r. The BFS-tree T has at most k− 1 leaves and hence at most k− 2 vertices with out-degreeat least 2 in T . Now, let P = p1p2 . . . pl be a path in T such that all verticesin V (P ) have out-degree 1 in T (P does not need to be a maximal path of T ).Let T1 be the subtree of T induced by the vertices reachable from r in T withoutusing vertices in P and let T2 be the subtree of T rooted at the child r2 of pl inT . Since T is a BFS-tree, it does not have any forward arcs, and thus plr2 is the

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only arc from P to T2. Thus all arcs originating in P and ending outside of Pmust have their endpoint in T1.

Lemma 8.2.9 Let D be a reduced instance, T be a BFS-tree rooted at r, andP = p1p2 . . . pl be a path in T such that all vertices in V (P ) have out-degree 1 inT . Let upi ∈ A(D), for some i between 1 and l, be an arc with u /∈ P . There isa path Pupi

from r to pi using the arc upi, such that V (Pupi) ∩ V (P ) ⊆ pi, pl.

Proof. Let T1 be the subtree of T induced by the vertices reachable from r inT without using vertices in P and let T2 be the subtree of T rooted at the childr2 of pl in T . If u ∈ V (T1) there is a path from r to u avoiding P . Appendingthe arc upi to this path yields the desired path Pupi

, so assume u ∈ V (T2). If allpaths from r to u use the arc pl−1pl then pl−1pl is an arc disconnecting pl andr2 from r, contradicting the fact that Reduction Rule 8.2.1.3 can not be applied.Let P ′ be a path from r to u not using the arc pl−1pl. Let x be the last vertexfrom T1 visited by P ′. Since P ′ avoids pl−1pl we know that P ′ does not visit anyvertices of P \ pl after x. We obtain the desired path Pupi

by taking the pathfrom r to x in T1 followed by the subpath of P ′ from x to u appended by the arcupi.

Corollary 8.2.10 Let D be a reduced no-instance, T be a BFS-tree rooted at rand P = p1p2 . . . pl be a path in T such that all vertices in V (P ) have out-degree1 in T . There are at most k vertices in P that are endpoints of arcs originatingoutside of P .

Proof. Let S be the set of vertices in P\pl that are endpoints of arcs originatingoutside of P . For the sake of contradiction suppose that there are at least k + 1vertices in P that are endpoints of arcs originating outside of P . Then |S| ≥ k. ByLemma 8.2.9 there exists a path from the root r to every vertex in S, that avoidsvertices of P \ pl as an intermediate vertex. Using these paths we can build anr-out-tree with every vertex in S as a leaf. This r-out-tree can be extended toa r-out-branching with at least k leaves by Proposition 8.2.3, contradicting thefact that D is a no-instance.

Lemma 8.2.11 Let D be a reduced no-instance, T be a BFS-tree rooted at r andP = p1p2 . . . pl be a path in T such that all vertices in V (P ) have out-degree 1 inT . There are at most 7(k − 1) vertices outside of P that are endpoints of arcsoriginating in P .

Proof. Let X be the set of vertices outside P which are out-neighbors of thevertices on P . Let P ′ be the path from r to p1 in T and r2 be the unique childof pl in T . First, observe that since there are no forward arcs, r2 is the only out-neighbor of vertices in V (P ) in the subtree of T rooted at r2. In order to boundthe size of X, we differentiate between two kinds of out-neighbors of vertices onP .

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• Out-neighbors of P that are not in V (P ′).

• Out-neighbors of P in V (P ′).

First, observe that |X \ V (P ′)| ≤ k − 1. Otherwise we could have made an r-out-tree with at least k leaves by taking the path P ′P and adding X \ V (P ′) asleaves with parents in V (P ).

In the rest of the proof we bound |X ∩V (P ′)|. Let Y be the set of vertices onP ′ with out-degree at least 2 in T and let P1, P2, . . . , Pt be the remaining subpathsof P ′ when vertices in Y are removed. For every i ≤ t, Pi = vi1vi2 . . . viq. Wedefine the vertex set Z to contain the two last vertices of each path Pi. Thenumber of vertices with out-degree at least 2 in T is upper bounded by k − 2 asT has at most k − 1 leaves. Hence, |Y | ≤ k − 2, t ≤ k − 1 and |Z| ≤ 2(k − 1).

Claim 8.2.12 For every path Pi = vi1vi2 . . . viq, 1 ≤ i ≤ t, there is either an arcuiviq−1 or uiviq where ui /∈ V (Pi).

To see the claim observe that the removal of arc viq−2viq−1 does not disconnectthe root r from both viq−1 and viq else Rule 8.2.1.3 would have been applicableto our reduced instance. For brevity assume that viq−1 is reachable from r afterthe removal of arc viq−2viq−1. Hence there exists a path from r to viq. Let uiviq

be the last arc of this path. The fact that the BFS-tree T does not have anyforward arcs implies that ui /∈ V (Pi).

To every path Pi = vi1vi2 . . . viq, 1 ≤ i ≤ t, we associate an interval Ii =vi1vi2 . . . viq−2 and an arc uiviq′, q

′ ∈ q − 1, q. This arc exists by Claim 8.2.12.Claim 8.2.12 and Lemma 8.2.9 together imply that for every path Pi there is apath Pri from the root r to viq′ that does not use any vertex in V (Pi)\viq−1, viqas an intermediate vertex. That is, V (Pri ∩ (V (Pi) \ viq−1, viq) = ∅.

Let P ′ri be a subpath of Pri starting at a vertex xi before vi1 on P ′ and ending

in a vertex yi after viq−2 on P ′. We say that a path P ′ri covers a vertex x if x is

on the subpath of P ′ between xi and yi and we say that it covers an interval Ijif xi appears before vj1 on the path P ′ and yi appears after vjq−2 on P ′. Observethat the path P ′

ri covers the interval Ii.Let P = P ′

1, P′2, . . . , P

′l ⊆ P ′

r1, . . . , P′rt be a minimum collection of paths,

such that every interval Ii, 1 ≤ i ≤ t, is covered by at least one of the pathsin P. Furthermore, let the paths of P be numbered by the appearance of theirfirst vertex on P ′. The minimality of P implies that for every P ′

i ∈ P there is aninterval I ′i ∈ I1, . . . , It such that P ′

i is the only path in P that covers I ′i.

Claim 8.2.13 For every 1 ≤ i ≤ l, no vertex of P ′ is covered by both P ′i and

P ′i+3.

The path P ′i+1 is the only path in P that covers the interval I ′i+1 and hence P ′

i

does not cover the last vertex of I ′i+1. Similarly P ′i+2 is the only path in P that

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covers the interval I ′i+2 and hence P ′i+3 does not cover the first vertex of I ′i+2.

Thus the set of vertices covered by both P ′i and P ′

i+3 is empty.Since paths P ′

i and P ′i+3 do not cover a common vertex, we have that the end

vertex of P ′i appears before the start vertex of P ′

i+3 on P ′ or is the same as thestart vertex of P ′

i+3. Partition the paths of P into three sets P0,P1,P2, wherepath P ′

i ∈ Pi mod 3. Also let Ii be the set of intervals covered by Pi. Observe thatevery interval Ij , 1 ≤ j ≤ t, is part of some Ii for i ∈ 0, 1, 2.

Let i ≤ 3 and consider an interval Ij ∈ Ii. There is a path Pj′ ∈ Pi thatcovers Ij such that both endpoints of Pj′ and none of the inner vertices of Pj′ lieon P ′. Furthermore for any pair of paths Pa, Pb ∈ Pi such that a < b, there is asubpath in P ′ from the endpoint of Pa to the starting point of Pb. Thus for everyi ≤ 3 there is a path P ∗

i from the root r to p1 which does not use any vertex ofthe intervals covered by the paths in Pi.

We now claim that the total number of vertices on intervals Ij, 1 ≤ j ≤ t,which are out-neighbors of vertices on V (P ) is bounded by 3(k− 1). If not, thenfor some i, the number of out-neighbors in Ii is at least k. Now we can make anr-out-tree with k leaves by taking any r-out-tree in D[V (P ∗

i )∪V (P )] and addingthe out-neighbors of the vertices on V (P ) in Ii as leaves with parents in V (P ).

Summing up the obtained upper bounds yields |X| ≤ (k− 1) + |r2|+ |Y |+|Z|+3(k− 1) ≤ (k− 1)+1+ (k− 2)+2(k− 1)+3(k− 1) = 7(k− 1), concludingthe proof.

Remark: Observe that the path P used in Lemmas 8.2.9 and 8.2.11 and Corol-lary 8.2.10 need not be a maximal path in T with its vertices having out-degreeone in T .

8.2.3 Bounding the Length of a Path

Now we bound the size of any maximal path with every vertex having out-degreeone in T and use the results proved here together with the ones in the precioussection to bound the size of any reduced no-instance of Rooted k-Leaf Out-Branching by O(k3). For a reduced instance D, a BFS-tree T of D rooted at r,let P = p1p2 . . . pl be a path in T such that all vertices in V (P ) have out-degree1 in T , and let S be the set of vertices in V (P ) \ pl with an in-arc from theoutside of P .

Definition 8.2.14 A subforest F = (V (P ), A(F )) of D[V (P )] is said to be anice forest of P if the following three properties are satisfied: (a) F is a forest ofdirected trees rooted at vertices in S; (b) If pipj ∈ A(F ) and i < j then pi hasout-degree at least 2 in F or pj has in-degree 1 in D; and (c) If pipj ∈ A(F ) andi > j then for all q > i, pqpj /∈ A(D).

In order to bound the size of a reduced no-instance D we are going to considera nice forest with the maximum number of leaves. However, in order to do this,

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we first need to show the existence of a nice forest of P .In the following discussion let D be a reduced no-instance, T be a BFS-tree

T of D rooted at r, P = p1p2 . . . pl be a path in T such that all vertices in V (P )have out-degree 1 in T and S be the set of vertices in V (P ) \ pl with an in-arcfrom the outside of P .

Lemma 8.2.15 There is a nice forest in P .

Proof. We define a subgraph F of D[V (P )] as follows. The vertex set of Fis V (P ) and an arc ptps is in A(F ) if ps 6∈ S and t is the largest number sothat ptps ∈ A(D). Notice that all arcs of F are covered by property (b) in thedefinition of a nice forest.

We prove that F is a forest. Suppose for contradiction that there is a cycleC in F . By definition of F every vertex has in-degree at most 1, C must be adirected cycle. Since every vertex in S has in-degree 0 in F , C ∩S = ∅. Considerthe highest numbered vertex pi on C. Since P has no forward arcs, pi−1 is thepredecessor of pi in C. The construction of F implies that there can not be an arcpqpi where q > i in A(D). Also, pi does not have any in-arcs from outside of P .Thus, pi−1 disconnects pi from the root. Hence, by Rule 8.2.1.2 pipi−1 6∈ A(D).Let pj be the predecessor of pi−1 in C. Then j < i− 1, since pipi−1 6∈ A(D) andpi is the highest numbered vertex in C. Hence j = i − 2. This contradicts thefact that D is a reduced instance since the arc pi−2pi−1 disconnects pi−1 and pi

from the root r implying that Rule 8.2.1.3 can be applied. Since F is a forest andsince every vertex in V (P ) except for vertices in S have in-degree 1 we concludethat F is a forest of directed trees rooted at vertices in S. Since F is a forest andP has no forward arcs, F is a nice forest.

For a nice forest F of P , we define the set of key vertices of F to be the setof vertices in S, the leaves of F , the vertices of F with out-degree at least 2 andthe set of vertices whose parent in F has out-degree at least 2.

Lemma 8.2.16 Let F be a nice forest of P . There are at most 5(k − 1) keyvertices of F .

Proof. By the proof of Corollary 8.2.10 there is an r-out-tree TS with (V (TS) ∩V (P )) ⊆ (S ∪ pl) and (A(TS) ∩ A(P )) = ∅, such that all vertices in S \ plare leaves of TS. We build an r-out-tree TF = (V (TS) ∪ V (P ), A(TS) ∪ A(F )).Notice that every leaf of F is a leaf of TF , except possibly for pl. Since D is ano-instance TF has at most k−1 leaves and k−2 vertices with out-degree at least2. Thus, F has at most k leaves and at most k − 2 vertices with out-degree atleast 2. Hence the number of vertices in F whose parent in F has out-degree atleast 2 is at most 2k − 2. Finally, by Corollary 8.2.10, |S| ≤ k. Adding up theseupper bounds yields that there are at most k− 1 + k− 2 + 2k− 2 + k = 5(k− 1)key vertices of F .

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We can now turn our attention to a nice forest F of P with the maximumnumber of leaves. Our goal is to show that if the key vertices of F are too spacedout on P then some of our reduction rules must apply. First, however, we needsome more observations about the interplay between P and F .

Observation 8.2.17 [Unique Path] For any two vertices pi, pj in V (P ) suchthat i < j, pipi+1 . . . pj is the only path from pi to pj in D[V (P )].

Proof. As T is a BFS-tree it has no forward arcs. So any vertex set X =p1, p2, . . . , pq with q < |V (P )|, the arc pqpq+1 is the only arc in D from a vertexin X to a vertex in V (P ) \X.

Corollary 8.2.18 No arc pipi+1 is a forward arc of F .

Proof. If pipi+1 is a forward arc of F then there is a path from pi to pi+1 inF . By Observation 8.2.17 pipi+1 is the unique path from pi to pi+1 in D[V (P )].Hence pipi+1 ∈ A(F ) contradicting the fact that it is a forward arc.

Observation 8.2.19 Let ptpj be an arc in A(F ) such that neither pt nor pj arekey vertices, and t ∈ j − 1, j + 1, . . . , l. Then for all q > t, pqpj 6∈ A(D).

Observation 8.2.19 follows directly from the definitions of a nice forest and keyvertices.

Observation 8.2.20 If neither pi nor pi+1 are key vertices, then either pipi+1 /∈A(F ) or pi+1pi+2 /∈ A(F ).

Proof. Assume for contradiction that pipi+1 ∈ A(F ) and pi+1pi+2 ∈ A(F ). Sinceneither pi nor pi+1 are key vertices, both pi+1 and pi+2 must have in-degree 1 inD. Then the arc pipi+1 disconnects both pi+1 and pi+2 from the root r and Rule8.2.1.3 can be applied, contradicting the fact that D is a reduced instance.

In the following discussion let F be a nice forest of P with the maximumnumber of leaves and let P ′ = pxpx+1 . . . py be a subpath of P containing no keyvertices, and additionally having the property that px−1px /∈ A(F ) and pypy+1 /∈A(F ).

Lemma 8.2.21 V (P ′) induces a directed path in F .

Proof. We first prove that for any arc pipi+1 ∈ A(P ′) such that pipi+1 /∈ A(F ),there is a path from pi+1 to pi in F . Suppose for contradiction that there is nopath from pi+1 to pi in F , and let x be the parent of pi+1 in F . Then pipi+1 is nota backward arc of F and hence F ′ = (F \xpi+1)∪pipi+1 is a forest of out-treesrooted at vertices in S. Also, since pi+1 is not a key vertex, x has out-degree 1 in

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F and thus x is a leaf in F ′. Since pi is not a leaf in F , F ′ has one more leaf thanF . Now, every vertex with out-degree at least 2 in F has out-degree at least 2 inF ′. Additionally, pi has out-degree 2 in F ′. Hence F ′ is a nice forest of P withmore leaves than F , contradicting the choice of F .

Now, notice that by Observation 8.2.17 any path in D[V (P )] from a vertexu ∈ V (P ′) to a vertex v ∈ V (P ′) that contains a vertex w /∈ V (P ′) must containeither the arc px−1px or the arc pypy+1. Since neither of those two arcs are arcsof F it follows that for any arc pipi+1 ∈ A(P ′) such that pipi+1 /∈ A(F ), there isa path from pi+1 to pi in F [V (P ′)]. Hence F [V (P ′)] is weakly connected, that is,the underlying undirected graph is connected. Since every vertex in V (P ′) hasin-degree 1 and out-degree 1 in F we conclude that F [V (P ′)] is a directed path.

In the following discussion let Q′ be the directed path F [V (P ′)].

Observation 8.2.22 For any pair of vertices pi, pj ∈ V (P ′) if i ≤ j − 2 then pj

appears before pi in Q′.

Proof. Suppose for contradiction that pi appears before pj in Q′. By Observation8.2.17 pipi+1pi+2 . . . pj is the unique path from pi to pj in D[V (P ′)]. This pathcontains both the arc pipi+1 and pi+1pi+2 contradicting Observation 8.2.20.

Lemma 8.2.23 All arcs of D[V (P ′)] are contained in A(P ′) ∪A(F ).

Proof. Since P has no forward arcs it is enough to prove that any arc pjpi ∈A(D[V (P ′)]) with i < j is an arc of F . Suppose this is not the case and let pq

be the parent of pi in F . We know that pi has in-degree at least 2 in D and alsosince pi is not a key vertex pq has in-degree one in F . Hence by definition of Fbeing a nice forest, we have that for every t > q, ptpi /∈ A(D). It follows thati < j < q. By Lemma 8.2.21 F [V (P ′)] is a directed path Q′ containing both pi

and pj. If pj appears after pi in Q′, Observation 8.2.22 implies that i = j− 1 andthat pj has in-degree 1 in D since F is a nice forest. Thus pi separates pj fromthe root and Rule 8.2.1.2 can be applied to pjpi contradicting the fact that D isa reduced instance. Hence pj appears before pi in Q′.

Since pj is an ancestor of pi in F and pq is the parent of pi in F , pj is anancestor of pq in F and hence pq ∈ V (Q′) = V (P ′). Now, pj comes before pq

in Q′ and j < q so Observation 8.2.22 implies that q = j + 1 and that pq hasin-degree 1 in D since F is a nice forest. Thus pj separates pq from the root r andboth pjpi and pqpi are arcs of D. Hence Rule 8.2.1.4 can be applied to removethe arc pqpi contradicting the fact that D is a reduced instance.

Lemma 8.2.24 If |P ′| ≥ 3 there are exactly 2 vertices in P ′ that are endpointsof arcs starting outside of P ′.

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Proof. By Observation 8.2.17, px−1px is the only arc between p1, p2, . . . , px−1and P ′. By Lemma 8.2.21, F [V (P ′)] is a directed path Q′. Let pq be the firstvertex onQ′ and notice that the parent of pq in F is outside of V (P ′). Observation8.2.22 implies that q ≥ y − 1. Hence pq and px are two distinct vertices that areendpoints of arcs starting outside of P ′. It remains to prove that they are theonly such vertices. Let pi be any vertex in P ′ \ px, pq. By Lemma 8.2.21 V (P ′)induces a directed path Q′ in F , and since pq is the first vertex of Q′, the parentof pi in F is in V (P ′). Observation 8.2.19 yields then that ptpi 6∈ A(D) for anyt > y.

Observation 8.2.25 Let Q′ = F [V (P ′)]. For any pair of vertices u, v such thatthere is a path Q′[uv] from u to v in Q′, Q′[uv] is the unique path from u to v inD[V (P ′)].

Proof. By Lemma 8.2.21 Q′ is a directed path f1f2 . . . f|P ′| and let Q′[f1fi] bethe path f1f2 . . . fi. We prove that for any i < |Q′|, fifi+1 is the only arc fromV (Q′[f1fi]) to V (Q′[fi+1f|P ′|]). By Lemma 8.2.23 all arcs of D[V (P ′)] are eitherarcs of P ′ or arcs of Q′. Since Q′ is a path, fifi+1 is the only arc from V (Q′[f1fi])to V (Q′[fi+1f|P ′|]) in Q′. By Corollary 8.2.18 there are no arcs from V (Q′[f1fi])to V (Q′[fi+1f|P ′|]) in P ′, except possibly for fifi+1.

Lemma 8.2.26 For any vertex x /∈ V (P ′) there are at most 2 vertices in P ′ witharcs to x.

Proof. Suppose there are 3 vertices pa, pb, pc in V (P ′) such that a < b < c andsuch that pax, pbx, pcx ∈ A(D). By Lemma 8.2.21 Q′ = F [V (P ′)] is a directedpath. If pa appears before pb in Q′ then Observation 8.2.22 implies that a+1 = band that pb has in-degree 1 in D. Then pa separates pb from the root and henceRule 8.2.1.4 can be applied to remove the arc pbx contradicting the fact that D isa reduced instance. Hence pb appears before pa in Q′. By an identical argumentpc appears before pb in Q′.

Let Pb be a path in D from the root to pb and let u be the last vertex in Pb

outside of V (P ′). Let v be the vertex in Pb after u. By Lemma 8.2.24, u is eitherpx or the first vertex pq of Q′. If u = px then Observation 8.2.17 implies that Pb

contains pa, whereas if u = pq then Observation 8.2.25 implies that Pb containspc. Thus the set pa, pc separates pb from the root and hence Rule 8.2.1.4 can beapplied to remove the arc pbx contradicting the fact that D is a reduced instance.

Corollary 8.2.27 There are at most 14(k− 1) vertices in P ′ with out-neighborsoutside of P ′.

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Proof. By Lemma 8.2.11 there are at most 7(k − 1) vertices that are endpointsof arcs originating in P ′. By Lemma 8.2.26 each such vertex is the endpoint ofat most 2 arcs from vertices in P ′.

Lemma 8.2.28 |P ′| ≤ 154(k − 1) + 10.

Proof. Assume for contradiction that |P ′| > 154(k − 1) + 10 and let X be theset of vertices in P ′ with arcs to vertices outside of P ′. By Corollary 8.2.27,|X| ≤ 14(k − 1). Hence there is a subpath of P ′ on at least (154(k − 1) +10)/(14(k − 1) + 1) = 9 vertices containing no vertices of X. By Observation8.2.20 there is a subpath P ′′ = papa+1 . . . pb of P ′ on 7 or 8 vertices such thatneither pa−1pa nor pbpb+1 are arcs of F . By Lemma 8.2.21 F [V (P ′′)] is a directedpath Q′′. Let pq and pt be the first and last vertices of Q′′, respectively. ByLemma 8.2.24 pa and pq are the only vertices with in-arcs from outside of P ′′.By Observation 8.2.22 pq ∈ pb−1, pb and pt ∈ pa, pa+1. By the choice of P ′′

no vertex of P ′′ has an arc to a vertex outside of P ′. Furthermore, since P ′′ isa subpath of P ′ and Q′′ is a subpath of Q′ Lemma 8.2.23 implies that pb and pt

are the only vertices of P ′ with out-arcs to the outside of P ′′. By Lemma 8.2.17,the path P ′′ is the unique out-branching of D[V (P ′′)] rooted at pa. By Lemma8.2.25, the path Q′′ is the unique out-branching of D[V (P ′′)] rooted at pq. ByObservation 8.2.22 pb−2 appears before pa+2 in Q′′ and hence the vertex after pb

in Q′′ and pt+1 is not the same vertex. Thus Rule 8.2.1.5 can be applied on P ′′,contradicting the fact that D is a reduced instance.

Lemma 8.2.29 Let D be a reduced no-instance to Rooted k-Leaf Out-Branching.Then |V (D)| = O(k3). More specifically, |V (D)| ≤ 1540k3.

Proof. Let T be a BFS-tree of D. T has at most k − 1 leaves and at mostk − 2 inner vertices with out-degree at least 2. The remaining vertices can bepartitioned into at most 2k−3 paths P1 . . . Pt with all vertices having out-degree1 in T . We prove that for every q ∈ 1, . . . , t, |Pq| = O(k2). Let F be a niceforest of Pq with the maximum number of leaves. By Lemma 8.2.16, F has atmost 5(k − 1) key vertices. Let pi and pj be consecutive key vertices of F onPq. By Observation 8.2.20, there is a path P ′ = pxpx+1 . . . py containing no keyvertices, with x ≤ i+1 and y ≥ j−1, such that neither px−1px nor pypy+1 are arcsof F . By Lemma 8.2.28 |P ′| ≤ 154(k−1)+10 so |Pq| ≤ (5(k−1)+1)(154(k−1)+10)+3(5(k−1)). Hence, |V (D)| ≤ 2k(5k(154(k−1)+10+3)) ≤ 1540k3 = O(k3).

Lemma 8.2.29 results in a cubic kernel for Rooted k-Leaf Out-Branchingas follows.

Theorem 8.2.30 Rooted k-Leaf Out-Branching and Rooted k-LeafOut-Tree admits a kernel of size O(k3).

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Proof. Let D be the reduced instance of Rooted k-Leaf Out-Branchingobtained in polynomial time using Lemma 8.2.8. If the size of D is more than1540k3 then return Yes. Else we have an instance of size bounded by O(k3). Thecorrectness of this step follows from Lemma 8.2.29 which shows that any reducedno-instance to Rooted k-Leaf Out-Branching has size bounded by O(k3).The result for Rooted k-Leaf Out-Tree follows similarly.

Chapter 9

Kernelization Lower Bounds

9.1 Polynomial Parameter Transformations with

Turing Kernels

In the previous chapter we gave a cubic kernel for Rooted k-Leaf Out-Branching, yielding a polynomial turing kernel for k-Leaf Out-Branching.It is natural to ask whether k-Leaf Out-Branching has a polynomial kernel.The answer to this question, somewhat surprisingly, is no, unless PH=Σ3

p. Toshow this we first show that k-Leaf Out-Tree admits no polynomial kernelunless PH=Σ3

p by giving a composition algorithm. Then we give a polynomial pa-rameter transformation from k-Leaf Out-Tree to k-Leaf Out-Branching.Our polynomial parameter transformation differs somewhat from the commonway a Karp reduction is done. In particular, to give the transformation we em-ploy the cubic kernel for Rooted k-Leaf Out-Branching that we proved inthe previous chapter together with the fact that NP-completeness guarantees theexistence of polynomial time reductions.

Theorem 9.1.1 k-Leaf Out-Tree has no polynomial kernel unless PH=Σ3p.

Proof. The problem is NP-complete [9]. We prove that it is compositional andthus, Theorem 5.1.6 will imply the statement of the theorem. A simple composi-tion algorithm for this problem is as follows. On input (D1, k), (D2, k), . . . , (Dt, k)output the instance (D, k) where D is the disjoint union of D1, . . . , Dt. Since anout-tree must be completely contained in a connected component of the underly-ing undirected graph of D, (D, k) is a yes-instance to k-Leaf Out-Tree if andonly if any out of (Di, k), 1 ≤ i ≤ t, is. This concludes the proof.

A willow graph [53]D = (V,A1∪A2) is a directed graph such thatD′ = (V,A1)is a directed path P = p1p2 . . . pn on all vertices of D and D′′ = (V,A2) is adirected acyclic graph with one vertex r of in-degree 0, such that every arc ofA2 is a backwards arc of P . p1 is called the bottom vertex of the willow, pn

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is called the top of the willow and P is called the stem. A nice willow graphD = (V,A1 ∪ A2) is a willow graph where pnpn−1 and pnpn−2 are arcs of D,neither pn−1 nor pn−2 are incident to any other arcs of A2 and D′′ = (V,A2) hasa pn-out-branching.

Observation 9.1.2 Let D = (V,A1 ∪ A2) be a nice willow graph. Every out-branching of D with the maximum number of leaves is rooted at the top vertexpn.

Proof. Let P = p1p2 . . . pn be the stem of D and suppose for contradiction thatthere is an out-branching T with the maximum number of leaves rooted at pi,i < n. Since D is a nice willow D′ = (V,A2) has a pn-out-branching T ′. Sinceevery arc of A2 is a back arc of P , T ′[vj : j ≥ i] is an pn-out-branching ofD[vj : j ≥ i]. Then T ′′ = (V, vxvy ∈ A(T ′) : y ≥ i ∪ vxvy ∈ A(T ) : y < i) isan out-branching of D. If i = n− 1 then pn is not a leaf of T since the only arcsgoing out of the set pn, pn−1 start in pn. Thus, in this case, all leaves of T areleaves of T ′′ and pn−1 is a leaf of T ′′ and not a leaf of T , contradicting the factthat T has the maximum number of leaves.

Lemma 9.1.3 k-Leaf Out-Tree in nice willow graphs is NP-complete underKarp reductions.

Proof. We reduce from the well known NP-complete Set Cover problem [72].A set cover of a universe U is a family F ′ of sets over U such that every elementof u appears in some set in F ′. In the Set Cover problem one is given afamily F = S1, S2, . . . Sm of sets over a universe U , |U | = n, together with anumber b ≤ m and asked whether there is a set cover F ′ ⊂ F with |F ′| ≤ bof U . In our reduction we will assume that every element of U is contained inat least one set in F . We will also assume that b ≤ m − 2. These assumptionsare safe because if either of them does not hold, the Set Cover instance can beresolved in polynomial time. From an instance of Set Cover we build a digraphD = (V,A1 ∪ A2) as follows. The vertex set V of D is a root r, vertices si foreach 1 ≤ i ≤ m representing the sets in F , vertices ei, 1 ≤ i ≤ n representingelements in U and finally 2 vertices p and p′.

The arc set A2 is as follows, there is an arc from r to each vertex si, 1 ≤ i ≤ mand there is an arc from a vertex si representing a set to a vertex ej representingan element if ej ∈ Si. Furthermore, rp and rp′ are arcs in A2. Finally, we letA1 = ei+1ei : 1 ≤ i < n ∪ si+1si : 1 ≤ i < m ∪ e1sm, s1p, pp

′, p′r. Thisconcludes the description of D. We now proceed to prove that there is a set coverF ′ ⊂ F with |F ′| ≤ b if and only if there is an out-branching in D with at leastn+m+ 2 − b leaves.

Suppose that there is a set cover F ′ ⊂ F with |F ′| ≤ b. We build a directedtree T rooted at r as follows. Every vertex si, 1 ≤ i ≤ m, p and p′ has r as their

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parent. For every element ej , 1 ≤ i ≤ n we chose the parent of ej to be si suchthat ej ∈ Si and Si ∈ F ′ and for every i′ < i either Si′ /∈ |F ′| or ej /∈ Si′ . Sincethe only inner nodes of T except for the root r are vertices representing sets inthe set cover, T is an out-branching of D with at least n+m+ 2 − b leaves.

In the other direction suppose that there is an out-branching T of D with atleast n + m + 2 − b leaves, and suppose that T has the most leaves out of allout-branchings of D. Since D is a nice willow with r as top vertex, Observation9.1.2 implies that T is an r-out-branching of D. Now, if there is an arc ei+1ei ∈A(T ) then let sj be a vertex such that ei ∈ Sj . Then T ′ = (T \ ei+1ei) ∪ sjei

is an r-out-branching of D with as many leaves as T . Hence, without loss ofgenerality, for every i between 1 and n, the parent of ei in T is some sj . LetF ′ = Si : si is an inner vertex of T. F ′ is a set cover of U with size at mostn+m+ 2 − (n+m+ 2 − b) = b, concluding the proof.

Theorem 9.1.4 k-Leaf Out-Branching does not admit a polynomial kernelunless PH=Σ3

p.

Proof. We give a polynomial parameter transformation from k-Leaf Out-Tree to k-Leaf Out-Branching. Let (D, k) be an instance to k-Leaf Out-Tree. For every vertex v ∈ V we make an instance (D, v, k) to Rooted k-LeafOut-Tree. Clearly, (D, k) is a yes-instance for k-Leaf Out-Tree if and onlyif (D, v, k) is a yes-instance to Rooted k-Leaf Out-Tree for some v ∈ V .By Theorem 8.2.30 Rooted k-Leaf Out-Tree has a O(k3) kernel, so we canapply the kernelization algorithm for Rooted k-Leaf Out-Tree separatelyto each of the n instances of Rooted k-Leaf Out-Tree to get n instances(D1, v1, k), (D2, v2, k), . . ., (Dn, vn, k) with |V (Di)| = O(k3) for each i ≤ n. ByLemma 9.1.3, k-Leaf Out-Branching in nice willow graphs is NP-completeunder Karp reductions, so we can reduce each instance (Di, vi, k) of Rootedk-Leaf Out-Tree to an instance (Wi, bi) of k-Leaf Out-Branching in nicewillow graphs in polynomial time in |Di|, and hence in polynomial time in k.Thus, in each such instance, bi ≤ (k + 1)c for some fixed constant c independentof both n and k. Let bmax = maxi≤n bi. Without loss of generality, bi = bmax forevery i. This assumption is safe because if it does not hold we can modify theinstance (Wi, bi) by replacing bi with bmax, subdividing the last arc of the stembmax − bi times and adding an edge from ri to each subdivision vertex.

From the instances (W1, bmax), . . ., (Wn, bmax) we build an instance (D′, bmax+1) of k-Leaf Out-Branching. Let ri and si be the top and bottom vertices ofWi, respectively. We build D′ simply by taking the disjoint union of the willowsgraphs W1,W2, . . . ,Wn and adding in an arc risi+1 for i < n and the arc rns1.Let C be the directed cycle in D obtained by taking the stem of D′ and addingthe arc rns1.

If for any i ≤ n, Wi has an out-branching with at least bmax leaves, then Wi

has an out-branching rooted at ri with at least bmax leaves. We can extend this

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to an out-branching of D′ with at least bmax + 1 leaves by following C from ri.In the other direction suppose D′ has an out-branching T with at least bmax + 1leaves. Let i be the integer such that the root r of T is in V (Wi). For any vertexv in V (D′) outside of V (Wi), the only path from r to v in D′ is the directedpath from r to v in C. Hence, T has at most 1 leaf outside of V (Wi). Thus,T [V (W1)] contains an out-tree with at least bmax leaves. Since bmax ≤ (k + 1)c

the reduction is a polynomial parameter transformation from k-Leaf Out-Treeto k-Leaf Out-Branching. By Theorem 9.1.1 and Proposition 5.2.2 k-LeafOut-Branching has no polynomial kernel unless PH=Σ3

p.

9.2 Explicit Identification in Composition Algo-

rithms

The kernelization lower bound for k-Leaf Out-Branching presented in theprevious section, together with a result of Bodlaender et al. [24] that the DisjointCycles problem does not admit a polynomial kernel were the first non-trivialapplications of the framework developed by Bodlaender et al. [20] and Fortnowand Santhanam [70]. In this section we give several non-trivial applications ofthis framework. In particular we describe a “cookbook” for showing kernelizationlower bounds using explicit identification. We then apply this cookbook to showthat a wide variety of problems do not admit polynomial kernels, resolving severalopen problems posed in the literature [10, 15, 78, 81, 111]. To show that a problemdoes not admit a polynomial size kernel we go through the following steps.

1. Find a suitable parameterization of the problem considered. Quite oftenparameterizations that impose extra structure make it easier to give a com-position algorithm.

2. Define a suitable colored version of the problem. This is in order to getmore control over how solutions to problem instances can look.

3. Show that the colored version of the problem is NP-complete.

4. Give a polynomial parameter transformation from the colored to the uncol-ored version. This will imply that if the uncolored version has a polynomialkernel then so does the colored version. Hence kernelization lower boundsfor the colored version directly transfer to the original problem.

5. Show that the colored version parameterized by k is solvable in time 2kc ·nO(1) for a fixed constant c.

6. Finally, show that the colored version is compositional and thus has nopolynomial kernel. To do so, proceed as follows.

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(a) If the number of instances in the input to the composition algorithmis at least 2kc

then running the parameterized algorithm on each in-stance takes time polynomial in input size. This automatically yieldsa composition algorithm.

(b) If the number of instances is less than 2kc

, every instance receives anunique identifier. Notice that in order to uniquely code the identi-fiers (ID) of all instances, kc bits per instance is sufficient. The IDsare coded either as an integer, or as a subset of a poly(k) sized set.

(c) Use the coding power provided by colors and IDs to complete thecomposition algorithm.

In the following sections we show how to apply this approach to show incom-pressibility and kernelization lower bounds for a variety of problems.

9.2.1 Steiner Tree, Variants of Vertex Cover, and BoundedRank Set Cover

The problems Steiner Tree, Connected Vertex Cover (ConVC), Ca-pacitated Vertex Cover (CapVC), and Bounded Rank Set Cover aredefined as follows. In Steiner Tree we are given a graph a graphG = (T∪N,E)and an integer k and asked for a vertex set N ′ ⊆ N of size at most k suchthat G[T ∪N ′] is connected. In ConVC we are given a graph G = (V,E) and aninteger k and asked for a vertex cover of size at most k that induces a connectedsubgraph in G. A vertex cover is a set C ⊆ V such that each edge in E has atleast one endpoint in C. The problem CapVC takes as input a graph G = (V,E),a capacity function cap : V → N

+ and an integer k, and the task is to find avertex cover C and a mapping from E to C in such a way that at most cap(v)edges are mapped to every vertex v ∈ C. Finally, an instance of BoundedRank Set Cover consists of a set family F over a universe U where everyset S ∈ F has size at most d, and a positive integer k. The task is to find asubfamily F ′ ⊆ F of size at most k such that ∪S∈F ′S = U . All four problemsare known to be NP-complete (e.g., see [72] and the proof of Theorem 9.2.3); inthis section, we show that the problems do not admit polynomial kernels for theparameter (|T |, k) (in the case of Steiner Tree), k (in the case of ConVCand CapVC), and (d, k) (in the case of Bounded Rank Set Cover), respec-tively. To this end, we first use the framework presented at the beginning of thischapter to prove that another problem, which is called RBDS, does not have apolynomial kernel. Then, by giving polynomial parameter transformations fromRBDS to the above problems, we show the non-existence of polynomial kernelsfor these problems.

In Red-Blue Dominating Set (RBDS) we are given a bipartite graphG = (T ∪ N,E) and an integer k and asked whether there exists a vertex

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set N ′ ⊆ N of size at most k such that every vertex in T has at least oneneighbor in N ′. We show that RBDS parameterized by (|T |, k) does not have apolynomial kernel. In the literature, the sets T and N are called “blue vertices”and “red vertices”, respectively. In this paper we will call the vertices “terminals”and “nonterminals” in order to avoid confusion with the colored version of theproblem that we are going to introduce. RBDS is equivalent to Set Cover andHitting Set and is, therefore, NP-complete [72].

In the colored version of RBDS, denoted by Colored Red-Blue Dom-inating Set (Col-RBDS), the vertices of N are colored with colors chosenfrom 1, . . . , k, that is, we are additionally given a function col : N → 1, . . . , k,and N ′ is required to contain exactly one vertex of each color. We will now followthe framework described at the beginning of this chapter.

Lemma 9.2.1 (1) The unparameterized version of Col-RBDS is NP-complete.(2) There is a polynomial parameter transformation from Col-RBDS to RBDS.(3) Col-RBDS is solvable in 2|T |+k · |T ∪N |O(1) time.

Proof. (1) It is easy to see that Col-RBDS is in NP. To prove its NP-hardness, we reduce the NP-complete problem RBDS to Col-RBDS: Givenan instance (G = (T ∪N,E), k) of RBDS, we construct an instance (G′ = (T ∪N ′, E ′), k, col) of Col-RBDS where the vertex setN ′ consists of k copies v1, . . . , vk

of every vertex v ∈ V , one copy of each color. That is, N ′ =⋃

a∈1,...,kva | v ∈N, and the color of every vertex va ∈ Na is col(va) = a. The edge set E ′ is givenby

E ′ =⋃

a∈1,...,ku, va | u ∈ T ∧ a ∈ 1, . . . , k ∧ u, v ∈ E .

Now, there is a set S ⊂ N of size k dominating all vertices in T in G if and onlyif in G′, there is a set S ′ ⊂ N ′ of size k containing one vertex of each color.

(2) Given an instance (G = (T ∪ N,E), k, col) of Col-RBDS, we constructan instance (G′ = (T ′ ∪ N,E ′), k) of RBDS. The set T ′ consists of all verticesfrom T plus k additional vertices z1, . . . , zk. The edge set E ′ consists of all edgesfrom E plus the edges

za, v | a ∈ 1, . . . , k ∧ v ∈ N ∧ col(v) = a.

Now, there is a set S ′ ⊂ N ′ of size k dominating all vertices in T ′ in G′ if andonly if in G, there is a set S ⊂ N of size k containing one vertex of each color.

(3) To solve Col-RBDS in the claimed running time, we first use the re-duction given in (2) from Col-RBDS to RBDS. The number |T ′| of termi-nals in the constructed instance of RBDS is |T | + k. Next, we transform theRBDS instance (G′, k) into an instance (F , U, k) of Set Cover where theelements in U one-to-one correspond to the vertices in T ′ and the sets in F

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one-to-one correspond to the vertices in N . Since Set Cover can be solvedin O(2|U | · |U | · |F|) time [68, Lemma 2], statement (3) follows.

Lemma 9.2.2 Col-RBDS parameterized by (|T |, k) is compositional.

Proof. Given a sequence

(G1 = (T1 ∪N1, E1), k, col1), . . . , (Gt = (Tt ∪Nt, Et), k, colt)

of Col-RBDS instances with |T1| = |T2| = . . . = |Tt| = p, we show how toconstruct a Col-RBDS instance (G = (T ∪N,E), k, col) as described in Defini-tion 5.1.5.

For i ∈ 1, . . . , t, let Ti := ui1, . . . , u

ip and Ni := vi

1, . . . , viqi. We start

with adding p vertices u1, . . . , up to the set T of terminals to be constructed. (Wewill add more vertices to T later.) Next, we add to the set N of nonterminalsall vertices from the vertex sets N1, . . . , Nt, preserving the colors of the vertices.That is, we set N =

⋃i∈1,...,tNi, and for every vertex vi

j ∈ N we define col(vij) =

coli(vij). Now, we add the edge set

⋃i∈1,...,t

uj1, v

ij2 | ui

j1, vij2 ∈ Ei

to G (see

Figure 9.1). The graphG and the coloring col constructed so far have the followingproperty: If at least one of the Col-RBDS instances (G1, k, col1), . . . , (Gt, k, colt)is a yes-instance, then (G, k, col) is also a yes-instance because if for any i ∈1, . . . , t a size-k subset fromNi dominates all vertices in Ti, then the same vertexset selected fromN also dominates all vertices in T . However, (G, k, col) may evenbe a yes-instance in the case where all instances (G1, k, col1), . . . , (Gt, k, colt) areno-instances, because in G one can select vertices into the solution that originatefrom different instances of the input sequence.

To ensure the correctness of the composition, we add more vertices andedges to G. We define for every graph Gi of the input sequence a uniqueidentifier ID(Gi), which consists of a size-(p + k) subset of 1, . . . , 2(p + k)Since

(2(p+k)

p+k

)≥ 2p+k and since we can assume that the input sequence does not

contain more than 2p+k instances, it is always possible to assign unique identi-fiers to all instances of the input sequence. (Note that if there are more than2p+k instances, then we can solve all these instances in

∑ti=1 2p+k · (p+ qi)

O(1) ≤t ·∑t

i=1(p + qi)O(1) time, which yields a composition algorithm.) For each color

pair (a, b) ∈ 1, . . . , k × 1, . . . , k with a 6= b, we add a vertex set W(a,b) =

w(a,b)1 , . . . , w

(a,b)2(p+k) to T , and we add to E the edge set

⋃

i∈1,...,t,j1∈1,...,qi

vi

j1, w

(a,b)j2

| a = col(vij1

) ∧ b ∈ 1, . . . , k \ a ∧ j2 ∈ ID(Gi)

∪

⋃

i∈1,...,t,j1∈1,...,qi

vi

j1 , w(a,b)j2

| b = col(vij1) ∧ a ∈ 1, . . . , k \ b ∧ j2 /∈ ID(Gi)

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N1

N1

T1

N2

N2

T2

W(white,black)

. . .. . .w

(white,black)1 . . . w

(white,black)4 . . . w

(white,black)7

ID(G1) =

1, 2, 5, 7ID(G2) =

2, 6, 7, 8

N

Figure 9.1: Example for the composition algorithm for Col-RBDS. The upperpart of the figure shows an input sequence consisting of two instances with k = 3(there are three colors: white, checkered, and black). The lower part of thefigure shows the output of the composition algorithm. For the sake of clarity,only the vertex set W(white,black) is displayed, whereas five other vertex sets W(a,b)

with a, b ∈ white, checkered, black are omitted. Since k = 3 and p = 5, eachID should consist of eight numbers, and W(white,black) should contain 16 vertices.For the sake of clarity, the displayed IDs consist of only four numbers each,and W(white,black) contains only eight vertices.

(see Figure 9.1).

Note that the construction conforms to the definition of a composition algo-rithm; in particular, k remains unchanged and the size of T is polynomial in p, kbecause |T | = p + k(k − 1) · 2(p + k). To prove the correctness of the construc-tion, we show that (G, k, col) has a solution N ′ ⊆ N if and only if at least oneinstance (Gi, k, coli) from the input sequence has a solution N ′

i ⊆ Ni.

In one direction, if N ′i ⊆ Ni is a solution for (Gi, k, coli), then the same vertex

set chosen from N forms a solution for (G, k, col). To see this, first note that thevertices from T are dominated by the chosen vertices. Moreover, for every colorpair (a, b) ∈ 1, . . . , k × 1, . . . , k with a 6= b, each vertex from W(a,b) is eitherconnected to all vertices v from Ni with col(v) = a or to all vertices v from Ni

with col(v) = b. Since N ′i contains one vertex of each color class from Ni, each

vertex in W(a,b) is dominated by a vertex from N chosen into the solution.

In the other direction, to show that any solution N ′ ⊆ N for (G, k, col) is a

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solution for at least one instance (Gi, k, coli), we prove that N ′ cannot containvertices originating from different instances of the input sequence. To this end,note that each two vertices in N ′ must have different colors. Assume, for the sakeof a contradiction, that N ′ contains a vertex vi1

j1with col(vi1

j1) = a originating

from the instance (Gi1 , k, coli1) and a vertex vi2j2

with col(vi2j2

) = b originatingfrom a different instance (Gi2 , k, coli2). Due to the construction of the IDs, wehave ID(Gi1) \ ID(Gi2) 6= ∅ and ID(Gi2) \ ID(Gi1) 6= ∅. This implies that there

are vertices in W(a,b) (namely, all vertices w(a,b)j with j ∈ ID(Gi2) \ ID(Gi1)) and

vertices in W(b,a) (namely, all vertices w(b,a)j with j ∈ ID(Gi1) \ ID(Gi2)) that are

neither adjacent to vi1j1

nor to vi2j2

. Therefore, N ′ does not dominate all verticesfrom T , which is a contradiction to the fact that N ′ is a solution for (G, k, col).

Theorem 9.2.3 Red-Blue Dominating Set and Steiner Tree, both pa-rameterized by (|T |, k), Connected Vertex Cover and Capacitated Ver-tex Cover, both parameterized by k, and Bounded Rank Set Cover, pa-rameterized by (k, d), do not admit polynomial kernels unless PH = Σ3

p.

Proof. For RBDS the statement of the theorem follows directly by Theorem5.1.6 together with Lemmata 9.2.1 and 9.2.2.

To show that the statement is true for the other four problems, we give poly-nomial parameter transformations from RBDS to each of them—due to Propo-sition 5.2.2, this suffices to prove the statement. Let (G = (T ∪ N,E), k) bean instance of RBDS. To transform it into an instance (G′ = (T ′ ∪ N,E ′), k)of Steiner Tree, define T ′ = T ∪ u where u is a new vertex and let E ′ =E ∪ u, vi | vi ∈ N. It is easy to see that every solution for Steiner Treeon (G′, k) one-to-one corresponds to a solution for RBDS on (G, k).

To transform (G, k) into an instance (G′′ = (V ′′, E ′′), k′′) of ConVC, firstconstruct the graph G′ = (T ′ ∪ N,E ′) as described above. The graph G′′ isthen obtained from G′ by attaching a leaf to every vertex in T ′. Now, G′′ has aconnected vertex cover of size k′′ = |T ′|+ k = |T |+ 1 + k iff G′ has a steiner treecontaining k vertices from N iff all vertices from T can be dominated in G byk vertices from N .

Next, we describe how to transform (G, k) into an instance (G′′′ = (V ′′′, E ′′′),cap, k′′′) of CapVC. First, for each vertex ui ∈ T , add a clique toG′′′ that containsfour vertices u0

i , u1i , u

2i , u

3i . Second, for each vertex vi ∈ N , add a vertex v′′′i

to G′′′. Finally, for each edge ui, vj ∈ E with ui ∈ T and vj ∈ N , add theedge u0

i , v′′′j toG′′′. The capacities of the vertices are defined as follows: For each

vertex ui ∈ T , the vertices u1i , u

2i , u

3i ∈ V ′′′ have capacity 1 and the vertex u0

i ∈ V ′′′

has capacity degG′′′(u0i )− 1. Each vertex v′′′i has capacity degG′′′(v′′′i ). Clearly, in

order to cover the edges of the size-4 cliques inserted for the vertices of T , everycapacitated vertex cover for G′′′ must contain all vertices u0

i , u1i , u

2i , u

3i . Moreover,

since the capacity of each vertex u0i is too small to cover all edges incident to u0

i ,

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at least one neighbor v′′′j of u0i must be selected into every capacitated vertex

cover for G′′′. Therefore, it is not hard to see that G′′′ has a capacitated vertexcover of size k′′′ = 4 · |T | + k iff all vertices from T can be dominated in G byk vertices from N .

Finally, to transform (G, k) into an instance (F , U, k) of Bounded Rank SetCover, add one element ei to U for every vertex ui ∈ T . For every vertex vj ∈ N ,add one set ei | ui, vj ∈ E to F . The correctness of the construction isobvious, and since |U | = |T |, every set in F contains at most d = |T | elements.

9.2.2 Unique Coverage

In the Unique Coverage problem we are given a universe U , a family of setsF over U and an integer k. The problem is to find a sub-family F ′ of F and a setS of elements in U such that |S| ≥ k and every element of S appears in exactlyone set in F ′, that is, the number of elements uniquely covered by F ′ is at leastk.

In order to obtain our negative results we have to utilize positive kernelizationresults for the problem. In some sense, we have to compress our instances as muchas possible in order to show that what remains is incompressible even though itis big. We utilize the following well-known and simple reduction rules for theproblem: (a) If any set S ∈ F contains at least k elements, return yes; (b) If anyelement e is not contained in any set in F , remove e from U ; and (c) If none ofthe above rules can be applied and |U | ≥ k(k − 1) return yes.

We show that the Unique Coverage problem does not have a polynomialkernel unless PH=Σ3

p. Notice that while the above reduction rules will compressthe instance to an instance with at most O(k2) elements, this is not a polynomialkernel because there is no polynomial bound on the size of F . Hence, our nega-tive result implies that unless PH=Σ3

p the size of F can not be compressed to apolynomial in k in polynomial time. We start by defining the colorful reduced ver-sion Colored Reduced Unique Coverage (Col-Red-UC) of the UniqueCoverage problem which is useful for making the composition algorithm. Inthis version the sets of F are colored with colors from the set 1, . . . , k and F ′ isrequired to contain exactly one set of each color. Furthermore, in Col-Red-UCevery set S in F has size at most k − 1 and |U | ≤ k2.

Lemma 9.2.4 (1) The unparameterized version of Col-Red-UC is NP-complete.(2) There is a polynomial parameter transformation from Col-Red-UC to UniqueCoverage. (3) Col-Red-UC parameterized by k is solvable in time O(k2k2

).

Proof. (1) To show that that Col-Red-UC is NP-complete we reduce from theUnique Coverage problem. For an instance (F , U, k) of Unique Coveragewe first apply the reduction rules above. We now assume that the given instance

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cannot be reduced any further. Now we make k copies of F , one copy of each color.This new instance has a colored subfamily uniquely covering at least k elementsif and only if (F , U, k) is a yes instance to Unique Coverage. Furthermore,even in the new instance we have that |U | ≤ k2 and that every set has at mostk − 1 elements.

(2) We now prove that there is a polynomial parameter transformation fromCol-Red-UC to Unique Coverage. For an instance (F , U, k) of Col-Red-UC we make a new instance (H, U ′, k′) to Unique Coverage. Let k′ = k(k2 +1) + k and for every color i we add a set Ui of k2 + 1 new elements to U andmake all sets colored with i contain Ui in addition to what they already contain.Thus we have that U ′ = U ∪⋃i∈1,...,kUi. Notice that in order to cover at least

k(k2 + 1) elements uniquely one has to pick exactly one set of each color. Thisconcludes the polynomial parameter transformation.

(3) Finally, observe that Col-Red-UC can be solved in time O(k2k2) because

the size of |U | is bounded by k2.

Lemma 9.2.5 The Col-Red-UC problem is compositional.

Proof. Given a sequence of Col-Red-UC instances I1 = (U,F1, k), . . . , It =(U,Ft, k), we construct an Col-Red-UC instance I = (U ′,F , k′). If the numberof instances t is at least 22k2 log k then running the algorithm from Lemma 9.2.4 onall instances takes time polynomial in the input size yielding a trivial compositionalgorithm. Thus we assume that t is at most 22k2 log k. We now construct ID’sfor for every instance, this is done in two steps. In the first step every instance igets a unique small id ID′(Ii) which is a subset of size k3/2 of the set 1, . . . , k3.The identifier of instance i is the set ID(Ii) which is defined to be ID(Ii) = x ∈N : ⌊x/k3⌋ ∈ ID′(Ii). In other words, ID(Ii) = k3 · j + j′ | j ∈ ID′(Ii) ∧ j′ ∈0, . . . , k3 − 1. Notice that the identifier of every instance is now a subset ofsize k6/2 of the set 1, . . . , k6 and that the ID’s of two different instances differin at least k3 places.

We start building the instance I by letting U ′ = U and F = F1 ∪F2 . . .∪Ft.The sets have the same color as in their respective instance. For every distinctordered pair of colors i, j ≤ k we add the set Ui,j = u1

i,j, . . . , uk6

i,j to U ′. For everyinstance Ip we consider the sets colored i and j respectively in Fp. To every setS with color i in Fp we add the set ux

i,j : x ∈ ID(Ip). Also, to every set S withcolor j in Fp we add the set ux

i,j : x /∈ ID(Ip). Finally we set k′ = k(k−1)k6+k.This concludes the construction.

If some Ip has a colored subfamily F ′ covering k elements uniquely, we showthat the same subfamily covers k′ elements uniquely in I. First note that F ′

covers k elements uniquely in U . It remains to prove that for every distinctordered pair i, j of colors, all elements of Ui,j are covered uniquely by F ′ in I.Consider an element uq

i,j ∈ Ui,j and let Si and Sj be the sets colored i and j

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respectively in F ′. If q ∈ ID(Ip) then Si contains uqi,j and Sj does not. Similarly

if q /∈ ID(Ip) then Sj contains uqi,j and Si does not. Furthermore no other set of

F ′ contains uqi,j and thus this element is uniquely covered.

In the other direction, suppose I has a colored subfamily F ′ covering k′ ele-ments uniquely. Suppose for contradiction that there is a color i and a color jsuch that the set Si ∈ F ′ with color i and the set Sj ∈ F ′ with color j originatefrom different instances. Observe that in the set Ui,j the sets Si and Sj intersectin at least k3 elements, and thus these elements are not covered uniquely by F ′.Then the total number of elements that can be uniquely covered by F ′ is upperbounded by k(k−1)k6 +k2−k3 < k(k−1)k6 < k′ yielding a contradiction. Thusall the sets in F ′ come from the same instance Ip and uniquely cover at least kelements in Ip. This concludes the proof.

Theorem 9.2.6 The Unique Coverage problem does not admit a polynomialkernel unless PH = Σ3

p.

9.2.3 Bounded Rank Disjoint Sets

In the Bounded Rank Disjoint Sets problem we are given a family F overa universe U with every set S ∈ F having size at most d together with a positiveinteger k. The question is whether there exists a subfamily F ′ of F with |F ′| ≥ ksuch that for every pair of sets S1, S2 ∈ F ′ we have that S1 ∩ S2 = ∅. Theproblem can be solved in time 2O(dk)nO(1) using color-coding and an applicationof dk-perfect hash families. We outline a proof which shows that this problemdoes not admit a poly(k, d) kernel. To do so we define a variation of the PerfectCode problem on graphs, which we call Bipartite Regular Perfect Codeproblem. In Bipartite Regular Perfect Code we are given a bipartitegraph G = (T ∪ N,E), where every vertex in N has the same degree, and aninteger k and asked whether there exists a vertex set N ′ ⊆ N of size at most ksuch that every vertex in T has exactly one neighbor in N ′. The set N ′ is calleda bipartite perfect code. Now we are ready to state the main theorem of thissubsection.

Theorem 9.2.7 Bipartite Regular Perfect Code parameterized by (|T |, k)and Bounded Rank Disjoint Sets parameterized by (d, k) do not have a poly-nomial kernel unless PH = Σ3

p.

Proof. We can show that Bipartite Regular Perfect Code parameter-ized by (|T |, k) does not have a polynomial kernel along the lines of the proofof Theorem 9.2.3, which shows that RBDS parameterized by (|T |, k) does nothave a polynomial kernel. Bipartite Regular Perfect Code is known tobe NP-complete even when every vertex in N has degree exactly 3 [99]. Theproof showing that the colored version of Bipartite Regular Perfect Code

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(1) is NP-complete, (2) has a fixed-parameter algorithm of the desired kind, and(3) has a polynomial parameter transformation to Bipartite Regular Per-fect Code is just a minor modification as for RBDS. For the composition,it is enough to observe that if the input graphs to the composition algorithmare one sided regular then the composed graph will remain one sided regular inLemma 9.2.2. This is true as every vertex is made adjacent to the same numberof newly added vertices and newly added vertices are added to T .

Finally, to show that Bounded Rank Disjoint Sets parameterized by(d, k) does not have a polynomial kernel we give a polynomial parameter transfor-mation from Bipartite Regular Perfect Code to Bounded Rank Dis-joint Sets. To this end, given an instance (G = (T ∪ N,E), k) for Bipar-tite Regular Perfect Code, we make an instance (U,F , k′, d) for BoundedRank Disjoint Sets as follows. Let U = T , F = Fv | v ∈ N, Fv = N(v),k′ = k and d = r where r is the degree of any vertex inN . Observe that k = |T |/r.From here it easily follows that G has a bipartite perfect code of size k if andonly if (U,F , k, d) has a subfamily F ′ of F of pairwise disjoint sets.

9.2.4 Domination and Transversals

In the Small Universe Hitting Set problem we are given a set family F overa universe U with |U | ≤ d together with a positive integer k. The question iswhether there exists a subset S in U of size at most k such that every set in F hasa non-empty intersection with S. We show that the Small Universe HittingSet problem parameterized by the solution size k and the size d = |U | of theuniverse does not have a kernel of size polynomial in (k, d) unless PH = Σ3

p.One should notice that while Hitting Set and Set Cover in fact are thesame problem, Small Universe Hitting Set and Small Universe SetCover are not, because in the former we are restricting the number of potentialdominators while in the later we restrict the number of objects to be dominated.

We define the colored version of Small Universe Hitting Set, called Col-SUHS as follows. We are given a set family F over a universe U with |U | ≤ d,and a positive integer k. The elements of U are colored with colors from the set1, . . . , k and the question is whether there exists a subset S ⊆ U containingexactly one element of each color such that every set in F has a non-emptyintersection with S.

Lemma 9.2.8 (1) The unparameterized version of Col-SUHS is NP-complete.(2) There is a polynomial parameter transformation from Col-SUHS to SmallUniverse Hitting Set. (3) Col-SUHS parameterized by d, k is solvable intime O(2d · nO(1)).

Proof. (1) We show that Col-SUHS is NP-complete by reducing from SmallUniverse Hitting Set, which is easily seen to be NP-complete by a reduction

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from Dominating Set. Given an instance (F , U, d, k) to Small UniverseHitting Set we make an instance (F ′, U ′, d · k, k) of Col-BRHS by letting U ′

contain k copies of U , with one element of each color. For every set S in F wemake a set S ′ in F ′ by letting S ′ contain all copies of all elements in S. From theconstruction it follows that (F , U, d, k) has a hitting set of size at most k if andonly if (F ′, U ′, d · k, k) has a colored hitting set of size k.

(2) We now give a polynomial parameter transformation from Col-SUHS toSmall Universe Hitting Set. For an instance (F ′, U, d, k) to Col-SUHS wemake an instance (F , U, d, k) to Small Universe Hitting Set as follows. Forevery i between 1 and k let Ui be the set of elements in U colored with i. Weconstruct F by starting with F ′ and for every i adding the set Ui to F ′. Now, asubset S of U hits every set of F if and only if it hits every set of F ′ and containsat least one vertex of each color.

(3) Finally, observe that Col-SUHS parameterized by d, k is solvable in timeO(2d · nO(1)) by enumerating all subsets of U and for each set checking whetherit is a hitting set with at least one vertex of each color.

Lemma 9.2.9 The problem Col-SUHS is compositional.

Proof. We have to show how, given a sequence of Col-SUHS instances (F1, U, d,k), . . ., (Ft, U, d, k) to Col-SUHS where |U | ≤ d, to construct a Col-SUHSinstance (F , U ′, d′, k′) as described in Definition 5.1.5.

If the number of instances is at least 2d then running the algorithm fromLemma 9.2.8 on all instances takes time polynomial in the input size yielding atrivial composition algorithm. Thus we can assume that t ≤ 2d. Furthermore,we need the number of instances to be a power of 2. To make this true we addin an appropriate number of no-instances. As we never add more than t extrainstances in order to make the number of instances a power of 2 this can be donein polynomial time and hence we can safely assume that t is a power of 2, say2l. Observe that since t ≤ 2d we have that l ≤ d. Now, let every instance beidentified by a unique number from 0 to t− 1.

We let k′ = k+l and start building (F , U ′, d′, k′) from (F1, U, d, k), . . . , (Ft, U, d,k) by letting U ′ = U and letting elements keep their color. For every i ≤ t we addthe family Fi to F . We now add 2l new elements C = a1, b1, . . . , al, bl to U ′ andfor every i ≤ l, ai, bi comprise a new color class. We conclude the constructionby modifying the sets in F that came from the input instances to the compositionalgorithm. For every j ≤ t we consider all sets in Fj. For every such set S weproceed as follows. Let ID(j) be the identification number of instance number j.For every i ≤ l we look at the i’th bit in the binary representation of ID(j). Ifthis bit is set to 1 we add ai to S and if the bit is set to 0 we add bi to S. Thisconcludes the construction.

Now, if there is a colored hitting set S for Fj with |S| ≤ k we construct acolored hitting set S ′ for F of size k+ l as follows. First we add S to S ′ and then

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we consider the identification number ID(j) of instance j. For every i between 1and l we consider the i’th bit of ID(j). If this bit is set to 1 we add bj to S ′ elsewe add aj to S ′. Clearly S ′ is a hitting set for Fi, has size k+ l and contains onevertex of each color. It remains to show that S ′ hits all other sets of F . Considerany set X ∈ Fq for some q 6= j. Then there is an i such that ID(q) differs fromID(j) in the i’th bit. If the i’th bit of ID(q) is 1 then the i’th bit of ID(j) is 0and hence ai ∈ S ′. Since the i’th bit of ID(q) is 1, ai ∈ X.

In the other direction, suppose there is a colored hitting set S ′ of size l + kof F . For every i ≤ l, exactly one out of the vertices ai and bi is in S ′. Let p bethe number between 0 and 2l − 1 such that for every i the i’th bit of p is 1 if andonly if bi ∈ S ′. Observe that the sets in F originating from the family Fj suchthat ID(j) = p do not contain any of the elements of S ′∩C. Thus S ′′ = S ′∩U isa colored hitting set for Fj containing at most one element from each color class.S ′′ can thus be extended to a colored hitting set S of Fj with |S| = k, concludingthe proof.

Theorem 9.2.10 Small Universe Hitting Set parameterized by solutionsize k and universe size |U | = d does not have a polynomial kernel unless PH =Σ3

p. The Dominating Set problem parameterized by the solution size k and thesize c of a minimum vertex cover of the input graph does not have a polynomialkernel.

Proof. The first part of the theorem follows immediately from Lemmata 9.2.8and 9.2.9. To show that the Dominating Set problem parameterized by thesolution size k and the size c of a minimum vertex cover of the input graph doesnot have a polynomial kernel we give a polynomial parameter transformationfrom Col-SUHS to Dominating Set. On input (F , U, d, k) to Col-SUHS wemake an instance (G, k) to Dominating Set as follows. We start by letting G bethe bipartite element-set incidence graph of (F , U). Now we and add one vertexvi for every color class i and make vi adjacent to every vertex in G correspondingto an element of U with color i. This concludes the construction.

First, observe that U is a vertex cover of size d of G. Second, observe that ifthere is a colored hitting set S with |S| ≤ k of (F , U, d, k) then S is a dominatingset of G. Finally, if S is a dominating set of G then for every i there is a vertex inS ∩N [vi]. Hence, S ∩ U is a hitting set for F containing at most one element ofeach color. Thus this set can be extended to a colorful hitting set of F , concludingthe proof.

Theorem 9.2.10 has some interesting consequences. For instance, it impliesthat the Hitting Set problem parameterized by solution size k and the max-imum size d of any set in F does not have a kernel of size poly(k, d) unlessPH = Σ3

p. The second part of Theorem 9.2.10 implies that the DominatingSet problem in graphs excluding a fixed graph H as a minor parameterized by

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(k, |H|) does not have a kernel of size poly(k, |H|) unless PH = Σ3p. This follows

from the well-known fact that every graph with a vertex cover of size c excludesthe complete graph Kc+2 as a minor. Similarly, since every graph with a vertexcover of size c is c-degenerate it follows that the Dominating Set problem ind-degenerate graphs does not have a kernel of size poly(k, d) unless PH = Σ3

p.

Theorem 9.2.11 Unless PH = Σ3p the problems Hitting Set parameterized

by solution size k and the maximum size d of any set in F , Dominating Setin H-Minor Free Graphs parameterized by (k, |H|), and Dominating Setparameterized by solution size k and degeneracy d of the input graph do not havea polynomial kernel.

9.2.5 Numeric Problem: Small Subset Sum

In the Subset Sum problem we are given a set S of n integers and a targett and asked whether there is a subset S ′ of S that adds up to exactly t. Inthe most common parameterization of this problem one is also given an integerk and asked whether there is a subset S ′ of S of size at most k that adds upto t. This parameterization, however, is W [1]-hard. We consider a strongerparameterization where in addition to k an extra parameter d is provided and theintegers in S are required to have size at most 2d. This version, Small SubsetSum is trivially fixed parameter tractable by dynamic programming. We believethat Small Subset Sum is the most restrictive plausible parameterization ofthe Subset Sum problem. We show that even this version does not admit apolynomial kernel, by giving a polynomial parameter transformation from theColored Red-Blue Dominating Set (Col-RBDS) problem.

Theorem 9.2.12 Small Subset Sum parameterized by (d, k) does not admita kernel polynomial in (d, k) unless PH = Σ3

p.

Proof. We give a polynomial parameter transformation from the ColoredRed-Blue Dominating Set (Col-RBDS) problem to Small Subset Sum.Given an instance (G = (T ∪N,E), k, d) to Col-RBDS, such that |T | = d andN has been colored with colors from 1, . . . , k, we build an instance (S, t, k′, d′)to Small Subset Sum. For an integer x ∈ S we treat x both as a number andas a string—the encoding of x in the number system with base k(k + 1).

We let t be a length-(d + 2k) string of digits representing the number 1 +k(k + 1)/2 and k′ = k(d + 1). Now, order the elements of T in some order, sayT = t1, t2, . . . , td. For a vertex v ∈ N we define the string Z(v) to be a string ond digits, where the i’th digit is set to 1 if v, ti ∈ E and 0 otherwise. For aninteger i between 1 and k we define the string B(i) to be a length-k string withzeroes everywhere, except in the i’th digit, which is 1 + k(k + 1)/2. For everyvertex x ∈ N we add a string to S: Let i be the color of x. We add the string

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B(i)Z(x)B(k + 1 − i) to S and we will say that this string (or the number itrepresents) corresponds to x. Even though the numbers in the Small SubsetSum are uncolored, we color the string corresponding to x with the same coloras x in order to ease the discussion. Finally we add a set of uncolored numbersto S: For every i between 1 and d and every j between 1 and k the string ui,j

is a string of length 2k + d with zeroes everywhere except for the k + i’th digit,which is set to j. This concludes the construction. One should notice that everynumber is bounded by (k(k + 1))2k+d and hence we can set d′ = 3(2k + d) log k.We prove that there is a set N ′ ⊆ N containing one vertex of each color suchthat every vertex of T has a neighbor in N ′ if and only if there is a set S ′ ⊆ S ofat most k′ numbers that add up to t.

Suppose that there is a set N ′ ⊆ N containing one vertex of each color suchthat every vertex of T has a neighbor in N ′. We pick S ′ ⊆ S as follows. Forevery vertex v ∈ N ′ we add the string corresponding to v to S ′. Furthermore, forevery i ≤ d, let xi be the number of neighbors the vertex ti ∈ T has in N ′. Weadd the set ui,j : 1 ≤ j ≤ k ∧ j 6= xi − 1 to S ′. Since we picked one number ofeach color, when we add up the numbers in S ′ there are no carries. Since everyvertex ti ∈ T sees at least one and at most k vertices in N ′, the k + i’th digit ofthe sum of all numbers in S ′ is exactly 1 + k(k + 1)/2.

In the other direction, suppose there is a set S ′ ⊆ S with at most k′ numbersthat add up to t. If S contains no numbers colored 1, then the last digit of∑

x∈S′ x must be zero, a contradiction. If S contains at least two numbers colored1 then

∑x∈S′ x > t again yielding a contradiction. Hence S ′ contains exactly one

number colored 1. Let s1 be the number in S ′ colored 1. Now, if S ′ contains nonumbers colored 2 then the second last digit of

∑x∈S′ x is zero and if S ′ contains

at least 2 numbers colored 2 then −s1 +∑

x∈S′ x > t − s1, again contradictingthat t =

∑x∈S′ x. Repeating the argument for the remaining colors yields that

S ′ contains exactly one number of each color. For every i ≤ k we let si be thenumber in S ′ colored i. For every i let vi be the vertex in N corresponding tosi. We prove that for every i the vertex ti ∈ T has a neighbor in N ′. To achievethis we argue that there is a number sj such that the k + i’th digit of sj is 1.Suppose for contradiction that this is not the case. Notice that since there isat most one number of each color in S ′ there are no carries when we add upthe numbers of S ′, even if S ′ contains all uncolored numbers. Hence, even ifS ′ contains all uncolored numbers whose k + i’th digit is non-zero the k + i’thdigit of

∑x∈S′ x must be strictly less than 1 + k(k + 1)/2, yielding the desired

contradiction, thereby completing the proof.

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Chapter 10

Concluding Remarks and OpenProblems

In this thesis, we have provided new methods for showing upper and lower boundsin Parameterized Algorithms and Complexity and in Kernelization. We concludethe thesis with a list of open problems that naturally arise from our results, orto which the techniques developed here might be amenable.

Algorithmic

• Our algorithm for k-Feedback Arc Set in Tournaments runs time 2O(√

k) +nO(1). Can the polylogarithmic term be removed from the exponent, thatis, can k-Feedback Arc Set in Tournaments be solved in time 2O(

√k)+nO(1)?

• Can the chromatic coding technique be applied to other problems in densestructures? In particular, in the Edge Bipartization problem input is agraph G and integer k. The objective is to find a set of at most k edgeswhose removal makes the graph bipartite. Is there an 2o(k) time algorithmfor the Edge Bipartization problem restricted to graphs where everyvertex has degree at least c · n for every fixed constant c > 0?

Hardness

• Is Max Cut FPT parameterized by cliquewidth? In this problem we aregiven a graph G of cliquewidth at most k and asked to color the vertices ofG black or white such that the maximum number of edges have endpointswith different colors. We conjecture that the problem is W[1]-hard.

• A well-known open problem is whether the Biclique problem is FPT. Inthis problem we are given a graph G and an integer k and asked whether Gcontains a complete bipartite graph with both bipartitions of size at least

143

144

k as subgraph. This problem is believed to be W[1]-hard. Can a novelapplication of explicit identification be used to prove that Biclique indeedis W[1]-hard?

Kernelization

• Do all compact p-min/eq/max-CMSO problems admit a linear kernel ingraphs of bounded genus? If so, do all quasi-compact p-min/eq/max-CMSO problems admit a linear kernel in graphs of bounded genus?

• Is there a semantical characterization of the problems that have finite inte-ger index?

• Can Theorems 8.1.1 and 8.1.3 be extended to graph classes excluding afixed graph H as a minor, or at least for minor closed graph classes withbounded local treewidth [66]?

• It does not seem possible to directly apply Theorems 8.1.1 and 8.1.3 to theOdd Cycle Transversal problem. Does Odd Cycle Transversaladmit a polynomial kernel in planar graphs?

• Longest Path is known not to admit a polynomial kernel unless PH=Σ3p [20],

even when input is restricted to planar graphs. Does Longest Path inplanar graphs admit a polynomial size turing kernel?

• Feedback Vertex Set is known to admit a O(k2) kernel [132]. DoesFeedback Vertex Set admit a linear size turing kernel?

Kernelization Lower Bounds

• Several of the problems we have considered are parameterized by two vari-ables, like Bounded Rank Hitting Set and Bounded Rank SetCover parameterized by solution size k and maximum set size d, andDominating Set in H-Minor Free Graphs parameterized by (k, |H|).While we show that the problems do not admit a kernel polynomial in bothk and d, the problems do admit a polynomial kernel in the solution sizek when the other parameter is regarded as a constant. In particular bothBounded Rank Hitting Set and Bounded Rank Set Cover admitkO(d) kernels while Dominating Set in H-Minor Free Graphs admitsa kf(|H|) kernel. Is there something in between? Or, more specifically, doBounded Rank Hitting Set and Bounded Rank Set Cover admitkernels of size f(d) · kO(1), and does Dominating Set in H-Minor FreeGraphs admit a f(|H|) · kO(1) kernel? We coin this kind of kernels, uni-formly polynomial kernels and ask the general question - which problemsadmit uniformly polynomial kernels, and which do not?

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• Is there a way to rule out the existence of polynomial size turing kernels foran FPT problem, up to some complexity-theoretical assumption?

146

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